DP Mathematics: Analysis and Approaches Questionbank
Topic 5 —Calculus
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Question
A curve has equation .
Find an expression for in terms of and .
Find the equations of the tangents to this curve at the points where the curve intersects the line .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to differentiate implicitly M1
A1A1A1
Note: Award A1 for correctly differentiating each term.
A1
Note: This final answer may be expressed in a number of different ways.
[5 marks]
A1
M1
at the tangent is and A1
at the tangent is A1
Note: These equations simplify to .
Note: Award A0M1A1A0 if just the positive value of is considered and just one tangent is found.
[4 marks]
Examiners report
Question
Consider the function defined by .
The curvature at any point on a graph is defined as .
Show that the function has a local maximum value when .
Find the -coordinate of the point of inflexion of the graph of .
Sketch the graph of , clearly indicating the position of the local maximum point, the point of inflexion and the axes intercepts.
Find the area of the region enclosed by the graph of and the -axis.
The curvature at any point on a graph is defined as .
Find the value of the curvature of the graph of at the local maximum point.
Find the value for and comment on its meaning with respect to the shape of the graph.
Markscheme
R1
R1
hence maximum at AG
[2 marks]
M1
A1
Note: Award M1A0 if extra zeros are seen.
[2 marks]

correct shape and correct domain A1
max at , point of inflexion at A1
zeros at and A1
Note: Penalize incorrect domain with first A mark; allow FT from (d) on extra points of inflexion.
[3 marks]
EITHER
M1A1
A1
OR
M1A1
A1
THEN
M1A1
A1
[6 marks]
(A1)
(A1)
A1
[3 marks]
A1
the graph is approximated by a straight line R1
[2 marks]
Examiners report
Question
A particle moves along a straight line. Its displacement, metres, at time seconds is given by . The first two times when the particle is at rest are denoted by and , where .
Find and .
Find the displacement of the particle when
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
M1A1
M1
A1A1
Note: Award A0A0 if answers are given in degrees.
[5 marks]
A1A1
[2 marks]
Examiners report
Question
Using the substitution show that .
Hence find the value of .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
let
(A1)
M1
Note: The method mark is for an attempt to substitute for both and .
(or equivalent) A1
when and when M1
AG
[4 marks]
M1
A1
A1
[3 marks]
Examiners report
Question
Find
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt at integration by parts with and M1
A1A1
Note: Award A1 for and A1 for .
solving by substitution with or inspection (M1)
A1
[5 marks]
Examiners report
Question
Consider the function defined by where is a positive constant.
The function is defined by for .
Showing any and intercepts, any maximum or minimum points and any asymptotes, sketch the following curves on separate axes.
;
Showing any and intercepts, any maximum or minimum points and any asymptotes, sketch the following curves on separate axes.
;
Showing any and intercepts, any maximum or minimum points and any asymptotes, sketch the following curves on separate axes.
.
Find .
By finding explain why is an increasing function.
Markscheme

A1 for correct shape
A1 for correct and intercepts and minimum point
[2 marks]

A1 for correct shape
A1 for correct vertical asymptotes
A1 for correct implied horizontal asymptote
A1 for correct maximum point
[??? marks]

A1 for reflecting negative branch from (ii) in the -axis
A1 for correctly labelled minimum point
[2 marks]
EITHER
attempt at integration by parts (M1)
A1A1
A1
A1
OR
attempt at integration by parts (M1)
A1A1
A1
A1
[5 marks]
M1A1A1
Note: Method mark is for differentiating the product. Award A1 for each correct term.
both parts of the expression are positive hence is positive R1
and therefore is an increasing function (for ) AG
[4 marks]
Examiners report
Question
Consider the function .
Express in the form .
Factorize .
Sketch the graph of , indicating on it the equations of the asymptotes, the coordinates of the -intercept and the local maximum.
Hence find the value of if .
Sketch the graph of .
Determine the area of the region enclosed between the graph of , the -axis and the lines with equations and .
Markscheme
A1
[1 mark]
A1
[1 mark]

A1 for the shape
A1 for the equation
A1 for asymptotes and
A1 for coordinates
A1 -intercept
[5 marks]
A1
M1
M1A1
[4 marks]

symmetry about the -axis M1
correct shape A1
Note: Allow FT from part (b).
[2 marks]
(M1)(A1)
A1
Note: Do not award FT from part (e).
[3 marks]
Examiners report
Question
Consider the function .
Show that the graph of is concave up for .
Sketch the graph of showing clearly any intercepts with the axes.
Markscheme
M1A1
for concave up R1AG
[3 marks]

-intercept at A1
-intercept at A1
stationary point of inflexion at with correct curvature either side A1
[3 marks]
Examiners report
Question
A particle moves in a straight line such that at time seconds , its velocity , in , is given by . Find the exact distance travelled by the particle in the first half-second.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt at integration by parts M1
A1
(A1)
Note: Condone absence of limits (or incorrect limits) and missing factor of 10 up to this point.
(M1)
A1
[5 marks]
Examiners report
Question
The folium of Descartes is a curve defined by the equation , shown in the following diagram.

Determine the exact coordinates of the point P on the curve where the tangent line is parallel to the -axis.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
M1A1
Note: Differentiation wrt is also acceptable.
(A1)
Note: All following marks may be awarded if the denominator is correct, but the numerator incorrect.
M1
EITHER
M1A1
A1
A1
OR
M1
A1
A1
A1
[8 marks]
Examiners report
Question
Consider the function .
Determine whether is an odd or even function, justifying your answer.
By using mathematical induction, prove that
where .
Hence or otherwise, find an expression for the derivative of with respect to .
Show that, for , the equation of the tangent to the curve at is .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
even function A1
since and is a product of even functions R1
OR
even function A1
since R1
Note: Do not award A0R1.
[2 marks]
consider the case
M1
hence true for R1
assume true for , ie, M1
Note: Do not award M1 for “let ” or “assume ” or equivalent.
consider :
(M1)
A1
A1
A1
so true and true true. Hence true for all R1
Note: To obtain the final R1, all the previous M marks must have been awarded.
[8 marks]
attempt to use (or correct product rule) M1
A1A1
Note: Award A1 for correct numerator and A1 for correct denominator.
[3 marks]
(M1)(A1)
(A1)
A1
A1
A1
Note: This A mark is independent from the previous marks.
M1A1
AG
[8 marks]
Examiners report
Question
Given that and , find
.
.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)(A1)
A1
A1
[4 marks]
(M1)
= 12 A1
[2 marks]
Examiners report
Question
Consider the functions defined for , given by and .
Hence, or otherwise, find .
Markscheme
METHOD 1
Attempt to add and (M1)
A1
(or equivalent) A1
Note: Condone absence of limits.
A1
METHOD 2
OR M1A1
A1
A1
[4 marks]
Examiners report
Question
Let
Find .
Find .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
M1A1
Note: M1 is for use of the chain rule.
[2 marks]
attempt at integration by parts M1
(A1)
A1
using integration by substitution or inspection (M1)
A1
Note: Award A1 for or equivalent.
Note: Condone lack of limits to this point.
attempt to substitute limits into their integral M1
A1
[7 marks]
Examiners report
Question
Use the substitution to find .
Hence find the value of , expressing your answer in the form arctan , where .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(accept or equivalent) A1
substitution, leading to an integrand in terms of M1
or equivalent A1
= 2 arctan A1
[4 marks]
= arctan 3 − arctan 1 A1
tan(arctan 3 − arctan 1) = (M1)
tan(arctan 3 − arctan 1) =
arctan 3 − arctan 1 = arctan A1
[3 marks]
Examiners report
Question
Let .
The graph of has a local maximum at A. Find the coordinates of A.
Show that there is exactly one point of inflexion, B, on the graph of .
The coordinates of B can be expressed in the form B where a, b. Find the value of a and the value of b.
Sketch the graph of showing clearly the position of the points A and B.
Markscheme
attempt to differentiate (M1)
A1
Note: Award M1 for using quotient or product rule award A1 if correct derivative seen even in unsimplified form, for example .
M1
A1
A1
[5 marks]
M1
A1
Note: Award A1 for correct derivative seen even if not simplified.
A1
hence (at most) one point of inflexion R1
Note: This mark is independent of the two A1 marks above. If they have shown or stated their equation has only one solution this mark can be awarded.
changes sign at R1
so exactly one point of inflexion
[5 marks]
A1
(M1)A1
Note: Award M1 for the substitution of their value for into .
[3 marks]
A1A1A1A1
A1 for shape for x < 0
A1 for shape for x > 0
A1 for maximum at A
A1 for POI at B.
Note: Only award last two A1s if A and B are placed in the correct quadrants, allowing for follow through.
[4 marks]
Examiners report
Question
Show that where .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
M1A1
AG
[2 marks]
METHOD 2
M1
A1
AG
[2 marks]
Examiners report
Question
Consider the curves and defined as follows
,
,
Using implicit differentiation, or otherwise, find for each curve in terms of and .
Let P(, ) be the unique point where the curves and intersect.
Show that the tangent to at P is perpendicular to the tangent to at P.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)
Note: M1 is for use of both product rule and implicit differentiation.
A1
Note: Accept
(M1)
A1
Note: Accept
[4 marks]
substituting and for and M1
product of gradients at P is or equivalent reasoning R1
Note: The R1 is dependent on the previous M1.
so tangents are perpendicular AG
[2 marks]
Examiners report
Question
The function is defined by , where 0 ≤ ≤ 5. The curve is shown on the following graph which has local maximum points at A and C and touches the -axis at B and D.
Use integration by parts to show that , .
Hence, show that , .
Find the -coordinates of A and of C , giving your answers in the form , where , .
Find the area enclosed by the curve and the -axis between B and D, as shaded on the diagram.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
attempt at integration by parts with , M1
A1
= M1A1
=
M1
AG
METHOD 2
attempt at integration by parts with , M1
A1
M1A1
M1
AG
METHOD 3
attempt at use of table M1
eg
A1A1
Note: A1 for first 2 lines correct, A1 for third line correct.
M1
M1
AG
[5 marks]
M1A1
A1
AG
Note: Do not accept solutions where the RHS is differentiated.
[3 marks]
M1A1
Note: Award M1 for an attempt at both the product rule and the chain rule.
(M1)
Note: Award M1 for an attempt to factorise or divide by .
discount (as this would also be a zero of the function)
(M1)
(at A) and (at C) A1A1
Note: Award A1 for each correct answer. If extra values are seen award A1A0.
[6 marks]
or A1
Note: The A1may be awarded for work seen in part (c).
M1
M1(A1)A1
Note: Award M1 for substitution of the end points and subtracting, (A1) for and and A1 for a completely correct answer.
[5 marks]
Examiners report
Question
Using the substitution , find .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)
valid attempt to write integral in terms of and M1
A1
(A1)
or equivalent A1
[5 marks]
Examiners report
Question
A camera at point C is 3 m from the edge of a straight section of road as shown in the following diagram. The camera detects a car travelling along the road at = 0. It then rotates, always pointing at the car, until the car passes O, the point on the edge of the road closest to the camera.
A car travels along the road at a speed of 24 ms−1. Let the position of the car be X and let OĈX = θ.
Find , the rate of rotation of the camera, in radians per second, at the instant the car passes the point O .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
let OX =
METHOD 1
(or −24) (A1)
(M1)
A1
EITHER
A1
attempt to substitute for into their differential equation M1
OR
A1
attempt to substitute for into their differential equation M1
THEN
(rad s−1) A1
Note: Accept −8 rad s−1.
METHOD 2
(or −24) (A1)
A1
attempt to differentiate implicitly with respect to M1
A1
attempt to substitute for into their differential equation M1
(rad s−1) A1
Note: Accept −8 rad s−1.
Note: Can be done by consideration of CX, use of Pythagoras.
METHOD 3
let the position of the car be at time be from O (A1)
M1
Note: For award A0M1 and follow through.
EITHER
attempt to differentiate implicitly with respect to M1
A1
attempt to substitute for into their differential equation M1
OR
M1
A1
at O, A1
THEN
A1
[6 marks]
Examiners report
Question
The curve is given by the equation .
At the point (1, 1) , show that .
Hence find the equation of the normal to at the point (1, 1).
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to differentiate implicitly M1
A1A1
Note: Award A1 for each term.
attempt to substitute , into their equation for M1
A1
AG
[5 marks]
attempt to use gradient of normal (M1)
so equation of normal is or A1
[2 marks]
Examiners report
Question
Find the coordinates of the points on the curve at which .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt at implicit differentiation M1
A1A1
Note: Award A1 for the second & third terms, A1 for the first term, fourth term & RHS equal to zero.
substitution of M1
A1
substitute either variable into original equation M1
(or ) A1
(or ) A1
, (3, −3) A1
[9 marks]
Examiners report
Question
A right circular cone of radius is inscribed in a sphere with centre O and radius as shown in the following diagram. The perpendicular height of the cone is , X denotes the centre of its base and B a point where the cone touches the sphere.
Show that the volume of the cone may be expressed by .
Given that there is one inscribed cone having a maximum volume, show that the volume of this cone is .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to use Pythagoras in triangle OXB M1
A1
substitution of their into formula for volume of cone M1
A1
Note: This A mark is independent and may be seen anywhere for the correct expansion of .
AG
[4 marks]
at max, R1
(since ) A1
EITHER
from part (a)
A1
A1
OR
A1
A1
THEN
AG
[4 marks]
Examiners report
Question
The graph of , 0 ≤ ≤ 5 is shown in the following diagram. The curve intercepts the -axis at (1, 0) and (4, 0) and has a local minimum at (3, −1).
The shaded area enclosed by the curve , the -axis and the -axis is 0.5. Given that ,
The area enclosed by the curve and the -axis between and is 2.5 .
Write down the -coordinate of the point of inflexion on the graph of .
find the value of .
find the value of .
Sketch the curve , 0 ≤ ≤ 5 indicating clearly the coordinates of the maximum and minimum points and any intercepts with the coordinate axes.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
3 A1
[1 mark]
attempt to use definite integral of (M1)
(A1)
= 3.5 A1
[3 marks]
(A1)
Note: (A1) is for −2.5.
= 1 A1
[2 marks]
A1A1A1
A1 for correct shape over approximately the correct domain
A1 for maximum and minimum (coordinates or horizontal lines from 3.5 and 1 are required),
A1 for -intercept at 3
[3 marks]
Examiners report
Question
Consider the functions and defined on the domain by and .
The following diagram shows the graphs of and
Find the -coordinates of the points of intersection of the two graphs.
Find the exact area of the shaded region, giving your answer in the form , where , .
At the points A and B on the diagram, the gradients of the two graphs are equal.
Determine the -coordinate of A on the graph of .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to form a quadratic in M1
A1
valid attempt to solve their quadratic M1
A1
A1A1
Note: Ignore any “extra” solutions.
[6 marks]
consider (±) M1
A1
Note: Ignore lack of or incorrect limits at this stage.
attempt to substitute their limits into their integral M1
A1A1
[5 marks]
attempt to differentiate both functions and equate M1
A1
attempt to solve for M1
A1
M1
A1
[6 marks]
Examiners report
Question
Show that .
Show that .
Hence or otherwise find in the form where , .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
M1A1
Note: Do not award the M1 for just .
Note: Do not award A1 if correct expression is followed by incorrect working.
AG
[2 marks]
M1
Note: M1 is for an attempt to change both terms into sine and cosine forms (with the same argument) or both terms into functions of .
A1A1
Note: Award A1 for numerator, A1 for denominator.
M1
AG
Note: Apply MS in reverse if candidates have worked from RHS to LHS.
Note: Alternative method using and in terms of .
[4 marks]
METHOD 1
A1
Note: Award A1 for correct expression with or without limits.
EITHER
or (M1)A1A1
Note: Award M1 for integration by inspection or substitution, A1 for , A1 for completely correct expression including limits.
M1
Note: Award M1 for substitution of limits into their integral and subtraction.
(A1)
OR
let M1
A1A1
Note: Award A1 for correct limits even if seen later, A1 for integral.
or A1
M1
THEN
Note: Award M1 for both putting the expression over a common denominator and for correct use of law of logarithms.
(M1)A1
METHOD 2
A1A1
A1A1(A1)
M1
M1A1
A1
[9 marks]
Examiners report
Question
Given that , find the value of .
Markscheme
seen (A1)
attempt at using limits in an integrated expression (M1)
(A1)
Setting their equation M1
Note: their equation must be an integrated expression with limits substituted.
A1
A1
Note: Do not award final A1 for .
[6 marks]
Examiners report
Question
Write in the form , where .
Hence, find the value of .
Markscheme
attempt to complete the square or multiplication and equating coefficients (M1)
A1
, ,
[2 marks]
use of their identity from part (a) (M1)
or A1
Note: Condone lack of, or incorrect limits up to this point.
(M1)
(A1)
A1
[5 marks]
Examiners report
Question
Consider .
For the graph of ,
Find .
Show that, if , then .
find the coordinates of the -intercept.
show that there are no -intercepts.
sketch the graph, showing clearly any asymptotic behaviour.
Show that .
The area enclosed by the graph of and the line can be expressed as . Find the value of .
Markscheme
attempt to use quotient rule (or equivalent) (M1)
A1
[2 marks]
simplifying numerator (may be seen in part (i)) (M1)
or equivalent quadratic equation A1
EITHER
use of quadratic formula
A1
OR
use of completing the square
A1
THEN
(since is outside the domain) AG
Note: Do not condone verification that .
Do not award the final A1 as follow through from part (i).
[3 marks]
(0, 4) A1
[1 mark]
A1
outside the domain R1
[2 marks]
A1A1
award A1 for concave up curve over correct domain with one minimum point in the first quadrant
award A1 for approaching asymptotically
[2 marks]
valid attempt to combine fractions (using common denominator) M1
A1
AG
[2 marks]
M1
( or) A1
area under the curve is M1
Note: Ignore absence of, or incorrect limits up to this point.
A1
A1
area is or M1
A1
[7 marks]
Examiners report
Question
Use l’Hôpital’s rule to determine the value of
.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
using l’Hôpital’s rule,
M1A1
(M1)A1
A1
A1
[6 marks]
Examiners report
Question
Find the equation of the tangent to the curve at the point where .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)
appreciate the need to find (M1)
A1
A1
A1
[5 marks]
Examiners report
Question
Consider the curve defined by .
Show that .
Prove that, when .
Hence find the coordinates of all points on , for , where .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt at implicit differentiation M1
A1M1A1
Note: Award A1 for LHS, M1 for attempt at chain rule, A1 for RHS.
M1
Note: Award M1 for collecting derivatives and factorising.
AG
[5 marks]
setting
(M1)
A1
OR OR A1
Note: If they offer values for , award A1 for at least two correct values in two different ‘quadrants’ and no incorrect values.
R1
A1
AG
[5 marks]
OR (M1)
A1A1
A1A1
Note: Allow ‘coordinates’ expressed as for example.
Note: Each of the A marks may be awarded independently and are not dependent on (M1) being awarded.
Note: Mark only the candidate’s first two attempts for each case of .
[5 marks]
Examiners report
Question
Consider the function defined by , where and .
Consider the case where .
State the equation of the vertical asymptote on the graph of .
State the equation of the horizontal asymptote on the graph of .
Use an algebraic method to determine whether is a self-inverse function.
Sketch the graph of , stating clearly the equations of any asymptotes and the coordinates of any points of intersections with the coordinate axes.
The region bounded by the -axis, the curve , and the lines and is rotated through about the -axis. Find the volume of the solid generated, giving your answer in the form , where .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
A1
[1 mark]
A1
[1 mark]
METHOD 1
M1
A1
A1
, (hence is self-inverse) R1
Note: The statement could be seen anywhere in the candidate’s working to award R1.
METHOD 2
M1
Note: Interchanging and can be done at any stage.
A1
A1
(hence is self-inverse) R1
[4 marks]
attempt to draw both branches of a rectangular hyperbola M1
and A1
and A1
[3 marks]
METHOD 1
(M1)
EITHER
attempt to express in the form M1
A1
OR
attempt to expand or and divide out M1
A1
THEN
A1
A1
A1
METHOD 2
(M1)
substituting A1
M1
A1
A1
Note: Ignore absence of or incorrect limits seen up to this point.
A1
[6 marks]
Examiners report
Question
Use l’Hôpital’s rule to find .
Markscheme
attempt to differentiate numerator and denominator M1
A1A1
Note: A1 for numerator and A1 for denominator. Do not condone absence of limits.
attempt to substitute (M1)
A1
Note: Award a maximum of M1A1A0M1A1 for absence of limits.
[5 marks]
Examiners report
Question
The acceleration, , of a particle moving in a horizontal line at time seconds, , is given by where is the particle’s velocity and .
At , the particle is at a fixed origin and has initial velocity .
Initially at , the particle moves in the positive direction until it reaches its maximum displacement from . The particle then returns to .
Let metres represent the particle’s displacement from and its maximum displacement from .
Let represent the particle’s velocity seconds before it reaches , where
.
Similarly, let represent the particle’s velocity seconds after it reaches .
By solving an appropriate differential equation, show that the particle’s velocity at time is given by .
Show that the time taken for the particle to reach satisfies the equation .
By solving an appropriate differential equation and using the result from part (b) (i), find an expression for in terms of .
By using the result to part (b) (i), show that .
Deduce a similar expression for in terms of .
Hence, show that .
Markscheme
(A1)
(or equivalent / use of integrating factor) M1
A1
EITHER
attempt to find with initial conditions M1
A1
A1
AG
OR
Attempt to find with initial conditions M1
A1
A1
AG
OR
A1
Attempt to find with initial conditions M1
A1
AG
Note: condone use of modulus within the ln function(s)
[6 marks]
recognition that when M1
A1
AG
Note: Award M1A0 for substituting into and showing that .
[6 marks]
(M1)
A1
( so) A1
at
Substituting into M1
A1
[5 marks]
METHOD 1
(M1)
A1
AG
METHOD 2
M1
A1
AG
[2 marks]
METHOD 1
(A1)
A1
METHOD 2
(A1)
A1
[2 marks]
METHOD 1
A1
attempt to express as a square M1
A1
so AG
METHOD 2
A1
Attempt to solve M1
minimum value of , (when ), hence R1
so AG
[3 marks]
Examiners report
Question
Let for .
Show that .
Use mathematical induction to prove that for .
Let .
Consider the function defined by for .
It is given that the term in the Maclaurin series for has a coefficient of .
Find the possible values of .
Markscheme
attempt to use the chain rule M1
A1
A1
AG
Note: Award M1A0A0 for or equivalent seen
[3 marks]
let
R1
Note: Award R0 for not starting at . Award subsequent marks as appropriate.
assume true for , (so ) M1
Note: Do not award M1 for statements such as “let ” or “ is true”. Subsequent marks can still be awarded.
consider
M1
(or equivalent) A1
EITHER
(or equivalent) A1
A1
Note: Award A1 for
A1
Note: Award A1 for leading coefficient of .
A1
OR
Note: The following A marks can be awarded in any order.
A1
Note: Award A1 for isolating correctly.
A1
Note: Award A1 for multiplying top and bottom by or .
A1
Note: Award A1 for leading coefficient of .
A1
THEN
since true for , and true for if true for , the statement is true for all, by mathematical induction R1
Note: To obtain the final R1, at least four of the previous marks must have been awarded.
[9 marks]
METHOD 1
using product rule to find (M1)
A1
A1
substituting into M1
A1
equating coefficient to M1
A1
or A1
METHOD 2
EITHER
attempt to find (M1)
A1
OR
attempt to apply binomial theorem for rational exponents (M1)
A1
THEN
(A1)
(M1)
coefficient of is A1
attempt to set equal to and solve M1
A1
or A1
METHOD 3
and (A1)
equating coefficient to M1
using product rule to find and (M1)
A1
substituting into M1
A1
A1
or A1
[8 marks]
Examiners report
Question
Solve the differential equation , given that at .
Give your answer in the form .
Markscheme
(M1)
attempt to find integrating factor (M1)
(A1)
attempt to use integration by parts (M1)
A1
substituting into an integrated equation involving M1
A1
[7 marks]
Examiners report
Question
Consider the expression where .
The binomial expansion of this expression, in ascending powers of , as far as the term in is , where .
Find the value of and the value of .
State the restriction which must be placed on for this expansion to be valid.
Markscheme
attempt to expand binomial with negative fractional power (M1)
A1
A1
attempt to equate coefficients of or (M1)
attempt to solve simultaneously (M1)
A1
[6 marks]
A1
[1 mark]
Examiners report
Question
Prove by mathematical induction that for .
Hence or otherwise, determine the Maclaurin series of in ascending powers of , up to and including the term in .
Hence or otherwise, determine the value of .
Markscheme
For
LHS: A1
RHS: A1
so true for
now assume true for ; i.e. M1
Note: Do not award M1 for statements such as "let ". Subsequent marks can still be awarded.
attempt to differentiate the RHS M1
A1
A1
so true for implies true for
therefore true and true true
therefore, true for all R1
Note: Award R1 only if three of the previous four marks have been awarded
[7 marks]
METHOD 1
attempt to use (M1)
Note: For , may be seen.
use of (M1)
A1
METHOD 2
' Maclaurin series of ' (M1)
(A1)
A1
[3 marks]
METHOD 1
attempt to substitute into M1
(A1)
EITHER
A1
OR
A1
THEN
so A1
METHOD 2
M1
(A1)
attempt to use L'Hôpital's rule M1
A1
[4 marks]
Examiners report
Question
Find the value of .
Markscheme
(A1)
A1A1
substituting limits into their integrated function and subtracting (M1)
OR
A1
[5 marks]
Examiners report
A mixed response was noted for this question. Candidates who simplified the algebraic fraction before integrating were far more successful in gaining full marks in this question. Many candidates used other valid approaches such as integration by substitution and integration by parts with varying degrees of success. A small number of candidates substituted the limits without integrating.
Question
A function is defined by where .
The graph of is shown below.
Show that is an odd function.
The range of is , where .
Find the value of and the value of .
Markscheme
attempts to replace with M1
A1
Note: Award M1A1 for an attempt to calculate both and independently, showing that they are equal.
Note: Award M1A0 for a graphical approach including evidence that either the graph is invariant after rotation by about the origin or the graph is invariant after a reflection in the -axis and then in the -axis (or vice versa).
so is an odd function AG
[2 marks]
attempts both product rule and chain rule differentiation to find M1
A1
sets their M1
A1
attempts to find at least one of (M1)
Note: Award M1 for an attempt to evaluate at least at one of their roots.
and A1
Note: Award A1 for .
[6 marks]
Examiners report
Question
By using the substitution or otherwise, find an expression for in terms of , where is a non-zero real number.
Markscheme
METHOD 1
(A1)
attempts to express the integral in terms of M1
A1
A1
Note: Condone the absence of or incorrect limits up to this point.
M1
A1
Note: Award M1 for correct substitution of their limits for into their antiderivative for (or given limits for into their antiderivative for ).
METHOD 2
(A1)
applies integration by inspection (M1)
A2
Note: Award A2 if the limits are not stated.
M1
Note: Award M1 for correct substitution into their antiderivative.
A1
[6 marks]
Examiners report
Question
The continuous random variable has probability density function
Find the value of .
Find .
Markscheme
attempt to integrate (M1)
A1
Note: Award (M1)A0 for .
Condone absence of up to this stage.
equating their integrand to M1
A1
[4 marks]
A1
Note: Condone absence of limits if seen at a later stage.
EITHER
attempt to integrate by inspection (M1)
A1
Note: Condone the use of up to this stage.
OR
for example,
Note: Other substitutions may be used. For example .
M1
Note: Condone absence of limits up to this stage.
A1
Note: Condone the use of up to this stage.
THEN
A1
Note: Award A0M1A1A0 for their or for working with incorrect or no limits.
[4 marks]
Examiners report
Most candidates who attempted part (a) knew that the integrand must be equated to 1 and only a small proportion of these managed to recognize the standard integral involved here. The effect of 3 in 3x2 was missed by many resulting in very few completely correct answers for this part. Part (b) proved to be challenging for vast majority of the candidates and was poorly done in general. Stronger candidates who made good progress in part (a) were often successful in part (b) as well. Most candidates used a substitution, however many struggled to make progress using this approach. Often when using a substitution, the limits were unchanged. If the function was re-written in terms of x, this did not result in an error in the final answer.
Question
A continuous random variable has the probability density function
.
The following diagram shows the graph of for .
Given that , find an expression for the median of in terms of and .
Markscheme
let be the median
EITHER
attempts to find the area of the required triangle M1
base is (A1)
and height is
area A1
OR
attempts to integrate the correct function M1
OR A1A1
Note: Award A1 for correct integration and A1 for correct limits.
THEN
sets up (their) or area M1
Note: Award M0A0A0M1A0A0 if candidates conclude that and set up their area or sum of integrals .
(A1)
as , rejects
so A1
[6 marks]
Examiners report
Question
A function is defined by , where .
A function is defined by , where .
The inverse of is .
A function is defined by , where .
Sketch the curve , clearly indicating any asymptotes with their equations. State the coordinates of any local maximum or minimum points and any points of intersection with the coordinate axes.
Show that .
State the domain of .
Given that , find the value of .
Give your answer in the form , where .
Markscheme
-intercept A1
Note: Accept an indication of on the -axis.
vertical asymptotes and A1
horizontal asymptote A1
uses a valid method to find the -coordinate of the local maximum point (M1)
Note: For example, uses the axis of symmetry or attempts to solve .
local maximum point A1
Note: Award (M1)A0 for a local maximum point at and coordinates not given.
three correct branches with correct asymptotic behaviour and the key features in approximately correct relative positions to each other A1
[6 marks]
M1
Note: Award M1 for interchanging and (this can be done at a later stage).
EITHER
attempts to complete the square M1
A1
A1
OR
attempts to solve for M1
A1
Note: Award A1 even if (in ) is missing
A1
THEN
A1
and hence is rejected R1
Note: Award R1 for concluding that the expression for must have the ‘’ sign.
The R1 may be awarded earlier for using the condition .
AG
[6 marks]
domain of is A1
[1 mark]
attempts to find (M1)
(A1)
attempts to solve for M1
EITHER
A1
attempts to find their M1
A1
Note: Award all available marks to this stage if is used instead of .
OR
A1
attempts to solve their quadratic equation M1
A1
Note: Award all available marks to this stage if is used instead of .
THEN
(as ) A1
Note: Award A1 for
[7 marks]
Examiners report
Part (a) was generally well done. It was pleasing to see how often candidates presented complete sketches here. Several decided to sketch using the reciprocal function. Occasionally, candidates omitted the upper branches or forgot to calculate the y-coordinate of the maximum.
Part (b): The majority of candidates knew how to start finding the inverse, and those who attempted completing the square or using the quadratic formula to solve for y made good progress (both methods equally seen). Otherwise, they got lost in the algebra. Very few explicitly justified the rejection of the negative root.
Part (c) was well done in general, with some algebraic errors seen in occasions.
Question
The function is defined by , where .
The function is defined by , where .
Find the Maclaurin series for up to and including the term.
Hence, find an approximate value for .
Show that satisfies the equation .
Hence, deduce that .
Using the result from part (c), find the Maclaurin series for up to and including the term.
Hence, or otherwise, determine the value of .
Markscheme
METHOD 1
recognition of both known series (M1)
and
attempt to multiply the two series up to and including term (M1)
(A1)
A1
METHOD 2
A1
and A1
substitute into or its derivatives to obtain Maclaurin series (M1)
A1
[4 marks]
(A1)
substituting their expression and attempt to integrate M1
Note: Condone absence of limits up to this stage.
A1
A1
[4 marks]
attempt to use product rule at least once M1
A1
A1
EITHER
A1
OR
A1
THEN
AG
Note: Accept working with each side separately to obtain .
[4 marks]
A1
AG
Note: Accept working with each side separately to obtain .
[1 mark]
attempt to substitute into a derivative (M1)
A1
(A1)
attempt to substitute into Maclaurin formula (M1)
A1
Note: Do not award any marks for approaches that do not use the part (c) result.
[5 marks]
METHOD 1
M1
(A1)
A1
Note: Condone the omission of in their working.
METHOD 2
indeterminate form, attempt to apply l'Hôpital's rule M1
, using l'Hôpital's rule again
, using l'Hôpital's rule again
A1
A1
[3 marks]
Examiners report
Part (a) was well answered using both methods. A number failed to see the connection between parts (a) and (b). Far too often, a candidate could not add three fractions together in part (b). There were many good responses to part (c) with candidates showing results on both sides are equal. A number of candidates failed to use the result from (c) in part (d). There were some good responses to part (e), with candidates working successfully with the series from (d) or applying l'Hôpital's rule. In particular, some responses were missing the appropriate limit notation and candidates following method 2 did not always show that the initial expression was of an indeterminate form before applying l'Hôpital's rule. Many candidates did not attempt parts (d) and (e).
Question
Use l’Hôpital’s rule to determine the value of .
Markscheme
* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
attempts to apply l’Hôpital’s rule on M1
M1A1A1
Note: Award M1 for attempting to use product and chain rule differentiation on the numerator, A1 for a correct numerator and A1 for a correct denominator. The awarding of A1 for the denominator is independent of the M1.
A1
[5 marks]
Examiners report
Question
A function is defined by .
The region is bounded by the curve , the -axis and the lines and . Let be the area of .
The line divides into two regions of equal area.
Let be the gradient of a tangent to the curve .
Sketch the curve , clearly indicating any asymptotes with their equations and stating the coordinates of any points of intersection with the axes.
Show that .
Find the value of .
Show that .
Show that the maximum value of is .
Markscheme
* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
a curve symmetrical about the -axis with correct concavity that has a local maximum point on the positive -axis A1
a curve clearly showing that as A1
A1
horizontal asymptote (-axis) A1
[4 marks]
attempts to find (M1)
A1
Note: Award M1A0 for obtaining where .
Note: Condone the absence of or use of incorrect limits to this stage.
(M1)
A1
AG
[4 marks]
METHOD 1
EITHER
(M1)
OR
(M1)
THEN
A1
A1
A1
METHOD 2
(M1)
A1
A1
A1
[4 marks]
attempts to find (M1)
A1
so AG
[2 marks]
attempts product rule or quotient rule differentiation M1
EITHER
A1
OR
A1
Note: Award A0 if the denominator is incorrect. Subsequent marks can be awarded.
THEN
attempts to express their as a rational fraction with a factorized numerator M1
attempts to solve their for M1
A1
from the curve, the maximum value of occurs at R1
(the minimum value of occurs at )
Note: Award R1 for any equivalent valid reasoning.
maximum value of is A1
leading to a maximum value of AG
[7 marks]
Examiners report
Question
The function is defined by .
Find the first two derivatives of and hence find the Maclaurin series for up to and including the term.
Show that the coefficient of in the Maclaurin series for is zero.
Using the Maclaurin series for and , find the Maclaurin series for up to and including the term.
Hence, or otherwise, find .
Markscheme
attempting to use the chain rule to find the first derivative M1
A1
attempting to use the product rule to find the second derivative M1
(or equivalent) A1
attempting to find , and M1
; ; A1
substitution into the Maclaurin formula M1
so the Maclaurin series for up to and including the term is A1
[8 marks]
METHOD 1
attempting to differentiate M1
(or equivalent) A2
substituting into their M1
so the coefficient of in the Maclaurin series for is zero AG
METHOD 2
substituting into the Maclaurin series for (M1)
substituting Maclaurin series for M1
A1
coefficient of is A1
so the coefficient of in the Maclaurin series for is zero AG
[4 marks]
substituting into the Maclaurin series for M1
A1
substituting into the Maclaurin series for M1
A1
selecting correct terms from above M1
A1
[6 marks]
METHOD 1
substitution of their series M1
A1
A1
METHOD 2
use of l’Hôpital’s rule M1
(or equivalent) A1
A1
[3 marks]
Examiners report
Question
An earth satellite moves in a path that can be described by the curve where and are in thousands of kilometres and is time in seconds.
Given that when , find the possible values of .
Give your answers in standard form.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
substituting for and attempting to solve for (or vice versa) (M1)
(A1)
EITHER
M1A1
OR
M1A1
THEN
attempting to find (M1)
A1
Note: Award all marks except the final A1 to candidates who do not consider ±.
METHOD 2
M1A1
(M1)(A1)
(M1)
A1
Note: Award all marks except the final A1 to candidates who do not consider ±.
[6 marks]
Examiners report
Question
The curve is defined by equation .
Find in terms of and .
Determine the equation of the tangent to at the point
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
M1A1A1
Note: Award A1 for the first two terms, A1 for the third term and the 0.
A1
Note: Accept .
Note: Accept .
[4 marks]
(M1)
(A1)
or equivalent A1
Note: Accept .
[3 marks]
Examiners report
Question
Consider the curve defined by the equation .
Find the volume of the solid formed when the region bounded by the curve, the -axis for and the -axis for is rotated through about the -axis.
Markscheme
Use of
(M1)(A1)
Note: Condone absence of limits or incorrect limits for M mark.
Do not condone absence of or multiples of .
A1
[3 marks]
Examiners report
Question
The region is enclosed by the graph of , the -axis and the line .
Write down a definite integral to represent the area of .
Calculate the area of .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
(M1)
(A1)
M1A1
Note: Award M1 for an attempt to find the difference between two functions, A1 for all correct.
METHOD 2
when A1
M1A1
Note: Award M1 for an attempt to find the inverse function.
A1
METHOD 3
M1A1A1A1
Note: Award M1 for considering the area below the -axis and above the -axis and A1 for each correct integral.
[4 marks]
A2
[2 marks]
Examiners report
Question
A water trough which is 10 metres long has a uniform cross-section in the shape of a semicircle with radius 0.5 metres. It is partly filled with water as shown in the following diagram of the cross-section. The centre of the circle is O and the angle KOL is radians.

The volume of water is increasing at a constant rate of .
Find an expression for the volume of water in the trough in terms of .
Calculate when .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
area of segment M1A1
A1
[3 marks]
METHOD 1
M1A1
(M1)
A1
METHOD 2
(M1)
A1
(M1)
A1
[4 marks]
Examiners report
Question
Xavier, the parachutist, jumps out of a plane at a height of metres above the ground. After free falling for 10 seconds his parachute opens. His velocity, , seconds after jumping from the plane, can be modelled by the function
His velocity when he reaches the ground is .
Find his velocity when .
Calculate the vertical distance Xavier travelled in the first 10 seconds.
Determine the value of .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)
A1
[2 marks]
(M1)
A1
[2 marks]
(M1)
A1
(M1)(A1)
A1
[5 marks]
Examiners report
Question
Consider
The function is defined by
The function is defined by .
Find the largest possible domain for to be a function.
Sketch the graph of showing clearly the equations of asymptotes and the coordinates of any intercepts with the axes.
Explain why is an even function.
Explain why the inverse function does not exist.
Find the inverse function and state its domain.
Find .
Hence, show that there are no solutions to ;
Hence, show that there are no solutions to .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)
or A1
[2 marks]

shape A1
and A1
-intercepts A1
[3 marks]
EITHER
is symmetrical about the -axis R1
OR
R1
[1 mark]
EITHER
is not one-to-one function R1
OR
horizontal line cuts twice R1
Note: Accept any equivalent correct statement.
[1 mark]
M1
M1
A1A1
[4 marks]
M1A1
A1
[3 marks]
M1
which is not in the domain of (hence no solutions to ) R1
[2 marks]
M1
as so no solutions to R1
Note: Accept: equation has no solutions.
[2 marks]
Examiners report
Question
By using the substitution , show that .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
EITHER
M1A1
M1A1
OR
M1A1
M1A1
THEN
(M1)
A1
M1
Note: This M1 may be seen anywhere, including a sketch of an appropriate triangle.
so AG
[7 marks]
Examiners report
Question
Consider the function .
Consider the region bounded by the curve , the -axis and the lines .
Show that the -coordinate of the minimum point on the curve satisfies the equation .
Determine the values of for which is a decreasing function.
Sketch the graph of showing clearly the minimum point and any asymptotic behaviour.
Find the coordinates of the point on the graph of where the normal to the graph is parallel to the line .
This region is now rotated through radians about the -axis. Find the volume of revolution.
Markscheme
attempt to use quotient rule or product rule M1
A1A1
Note: Award A1 for or equivalent and A1 for or equivalent.
setting M1
or equivalent A1
AG
[5 marks]
A1A1
Note: Award A1 for and A1 for . Accept .
[2 marks]

concave up curve over correct domain with one minimum point above the -axis. A1
approaches asymptotically A1
approaches asymptotically A1
Note: For the final A1 an asymptote must be seen, and must be seen on the -axis or in an equation.
[3 marks]
(A1)
attempt to solve for (M1)
A1
A1
[4 marks]
(M1)(A1)
Note: M1 is for an integral of the correct squared function (with or without limits and/or ).
A1
[3 marks]
Examiners report
Question
Consider the function .
Let .
Determine an expression for in terms of .
Sketch a graph of for .
Find the -coordinate(s) of the point(s) of inflexion of the graph of , labelling these clearly on the graph of .
Express in terms of .
Express in terms of .
Hence show that can be expressed as .
Solve the equation , giving your answers in the form where .
Markscheme
(or equivalent) (M1)A1
[2 marks]
A1A1A1A1
Note: Award A1 for correct behaviour at , A1 for correct domain and correct behaviour for , A1 for two clear intersections with -axis and minimum point, A1 for clear maximum point.
[4 marks]
A1
A1
[2 marks]
attempt to write in terms of only (M1)
A1
[2 marks]
(A1)
attempt to use (M1)
A1
[3 marks]
M1
(or equivalent) A1
AG
[2 marks]
or (M1)
A1
A1
Note: Only accept answers given the required form.
[3 marks]
Examiners report
Question
Given that can be expressed in the form , find the values of the constants , and .
Hence find .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
A1
A1
[2 marks]
M1M1
Note: Award M1 for dividing by to get , M1 for separating the and 1.
(M1)A1A1
Note: Award (M1)A1 for integrating , A1 for the other two terms.
[5 marks]
Examiners report
Question
A point P moves in a straight line with velocity ms−1 given by at time t seconds, where t ≥ 0.
Determine the first time t1 at which P has zero velocity.
Find an expression for the acceleration of P at time t.
Find the value of the acceleration of P at time t1.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to solve for t or equivalent (M1)
t1 = 0.441(s) A1
[2 marks]
M1A1
Note: Award M1 for attempting to differentiate using the product rule.
[2 marks]
(ms−2) A1
[1 mark]
Examiners report
Question
The following graph shows the two parts of the curve defined by the equation , and the normal to the curve at the point P(2 , 1).
Show that there are exactly two points on the curve where the gradient is zero.
Find the equation of the normal to the curve at the point P.
The normal at P cuts the curve again at the point Q. Find the -coordinate of Q.
The shaded region is rotated by 2 about the -axis. Find the volume of the solid formed.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
differentiating implicitly: M1
A1A1
Note: Award A1 for each side.
if then either or M1A1
two solutions for R1
not possible (as 0 ≠ 5) R1
hence exactly two points AG
Note: For a solution that only refers to the graph giving two solutions at and no solutions for award R1 only.
[7 marks]
at (2, 1) M1
(A1)
gradient of normal is 2 M1
1 = 4 + c (M1)
equation of normal is A1
[5 marks]
substituting (M1)
or (A1)
A1
[3 marks]
recognition of two volumes (M1)
volume M1A1A1
Note: Award M1 for attempt to use , A1 for limits, A1 for Condone omission of at this stage.
volume 2
EITHER
(M1)(A1)
OR
(M1)(A1)
THEN
total volume = 19.9 A1
[7 marks]
Examiners report
Question
A curve C is given by the implicit equation .
The curve intersects C at P and Q.
Show that .
Find the coordinates of P and Q.
Given that the gradients of the tangents to C at P and Q are m1 and m2 respectively, show that m1 × m2 = 1.
Find the coordinates of the three points on C, nearest the origin, where the tangent is parallel to the line .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt at implicit differentiation M1
A1M1A1
Note: Award A1 for first two terms. Award M1 for an attempt at chain rule A1 for last term.
A1
AG
[5 marks]
EITHER
when M1
(A1)
OR
or equivalent M1
(A1)
THEN
therefore A1
or A1
[4 marks]
m1 = M1A1
m2 = A1
m1 m2 = 1 AG
Note: Award M1A0A0 if decimal approximations are used.
Note: No FT applies.
[3 marks]
equate derivative to −1 M1
(A1)
R1
in the first case, attempt to solve M1
(0.486,0.486) A1
in the second case, and (M1)
(0,1), (1,0) A1
[7 marks]
Examiners report
Question
A function satisfies the conditions , and its second derivative is , ≥ 0.
Find .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)A1A1
Note: A1 for first term, A1 for second term. Withhold one A1 if extra terms are seen.
A1
Note: Allow FT from incorrect if it is of the form .
Accept .
attempt to use at least one boundary condition in their (M1)
,
⇒ A1
,
⇒
⇒ A1
[7 marks]
Examiners report
Question
Differentiate from first principles the function .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
M1
(A1)
A1
cancelling M1
then
A1
Note: Final A1 dependent on all previous marks.
METHOD 2
M1
(A1)
A1
cancelling M1
then
A1
Note: Final A1 dependent on all previous marks.
[5 marks]
Examiners report
Question
The function is defined by , 0 < < 3.
Draw a set of axes showing and values between −3 and 3. On these axes
Hence, or otherwise, find the coordinates of the point of inflexion on the graph of .
sketch the graph of , showing clearly any axis intercepts and giving the equations of any asymptotes.
sketch the graph of , showing clearly any axis intercepts and giving the equations of any asymptotes.
Hence, or otherwise, solve the inequality .
Markscheme
finding turning point of or finding root of (M1)
A1
(M1)A1
(0.899, −0.375)
Note: Do not accept . Accept y-coordinates rounding to −0.37 or −0.375 but not −0.38.
[4 marks]
smooth curve over the correct domain which does not cross the y-axis
and is concave down for > 1 A1
-intercept at 0.607 A1
equations of asymptotes given as = 0 and = 3 (the latter must be drawn) A1A1
[4 marks]
attempt to reflect graph of in = (M1)
smooth curve over the correct domain which does not cross the -axis and is concave down for > 1 A1
-intercept at 0.607 A1
equations of asymptotes given as = 0 and = 3 (the latter must be drawn) A1
Note: For FT from (i) to (ii) award max M1A0A1A0.
[4 marks]
solve or to get = 0.372 (M1)A1
0 < < 0.372 A1
Note: Do not award FT marks.
[3 marks]
Examiners report
Question
Let be the tangent to the curve at the point (1, ).
Find the coordinates of the point where meets the -axis.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
equation of tangent is OR (M1)(A1)
meets the -axis when
meets -axis at (0.667, 0) A1A1
Note: Award A1 for or seen and A1 for coordinates (, 0) given.
METHOD 1
Attempt to differentiate (M1)
when , (M1)
equation of the tangent is
meets -axis at
A1A1
Note: Award A1 for or seen and A1 for coordinates (, 0) given.
[4 marks]
Examiners report
Question
The function is defined by , .
Write down the range of .
Find , stating its domain.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
≥ 3 A1
[1 mark]
(M1)
Note: Exchange of variables can take place at any point.
(A1)
, ≥ 3 A1A1
Note: Allow follow through from (a) for last A1 mark which is independent of earlier marks in (b).
[4 marks]
Examiners report
Question
The function is defined by , .
Use integration by parts to find .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
write as (M1)
M1A1
(M1)(A1)
A1
METHOD 2
let M1
A1
M1
A1
M1A1
METHOD 3
Setting up and M1
M1A1
M1A1
A1
[6 marks]
Examiners report
Question
A particle moves along a horizontal line such that at time seconds, ≥ 0, its acceleration is given by = 2 − 1. When = 6 , its displacement from a fixed origin O is 18.25 m. When = 15, its displacement from O is 922.75 m. Find an expression for in terms of .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to integrate to find M1
A1
A1
attempt at substitution of given values (M1)
at
at
solve simultaneously: (M1)
A1
[6 marks]
Examiners report
Question
The function is defined by , ≥ 1 and the function is defined by , ≥ 0.
The region is bounded by the curves , and the lines , and as shown on the following diagram.
The shape of a clay vase can be modelled by rotating the region through 360˚ about the -axis.
Find the volume of clay used to make the vase.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
volume (M1)(M1)(M1)(A1)(A1)
Note: Award (M1) for use of formula for rotating about -axis, (M1) for finding at least one inverse, (M1) for subtracting volumes, (A1)(A1)for each correct expression, including limits.
A2
[7 marks]
Examiners report
Question
Consider the polynomial .
Sketch the graph of , stating clearly the coordinates of any maximum and minimum points and intersections with axes.
Hence, or otherwise, state the condition on such that all roots of the equation are real.
Markscheme
shape A1
-axis intercepts at (−3, 0), (1, 0) and -axis intercept at (0, −51) A1A1
minimum points at (−1.62, −118) and (3.72, 19.7) A1A1
maximum point at (2.40, 26.9) A1
Note: Coordinates may be seen on the graph or elsewhere.
Note: Accept −3, 1 and −51 marked on the axes.
[6 marks]
from graph, 19.7 ≤ ≤ 26.9 A1A1
Note: Award A1 for correct endpoints and A1 for correct inequalities.
[2 marks]
Examiners report
Question
The voltage in a circuit is given by the equation
, where is measured in seconds.
The current in this circuit is given by the equation
.
The power in this circuit is given by .
The average power in this circuit from to is given by the equation
, where .
Write down the maximum and minimum value of .
Write down two transformations that will transform the graph of onto the graph of .
Sketch the graph of for 0 ≤ ≤ 0.02 , showing clearly the coordinates of the first maximum and the first minimum.
Find the total time in the interval 0 ≤ ≤ 0.02 for which ≥ 3.
Find (0.007).
With reference to your graph of explain why > 0 for all > 0.
Given that can be written as where , , , > 0, use your graph to find the values of , , and .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
3, −3 A1A1
[2 marks]
stretch parallel to the -axis (with -axis invariant), scale factor A1
translation of (shift to the left by 0.003) A1
Note: Can be done in either order.
[2 marks]
correct shape over correct domain with correct endpoints A1
first maximum at (0.0035, 4.76) A1
first minimum at (0.0085, −1.24) A1
[3 marks]
≥ 3 between = 0.0016762 and 0.0053238 and = 0.011676 and 0.015324 (M1)(A1)
Note: Award M1A1 for either interval.
= 0.00730 A1
[3 marks]
(M1)
= 2.87 A1
[2 marks]
in each cycle the area under the axis is smaller than area above the axis R1
the curve begins with the positive part of the cycle R1
[2 marks]
(M1)
A1
A1
A1
(M1)
A1
[6 marks]
Examiners report
Question
A body moves in a straight line such that its velocity, , after seconds is given by for .
The following diagram shows the graph of against . Point is a local maximum and point is a local minimum.
The body first comes to rest at time . Find
Determine the coordinates of point and the coordinates of point .
Hence, write down the maximum speed of the body.
the value of .
the distance travelled between and .
the acceleration when .
Find the distance travelled in the first 30 seconds.
Markscheme
and A1A1A1A1
[4 marks]
maximum speed is A1
[1 mark]
(M1)A1
[2 marks]
(M1)
A1
[2 marks]
at (M1)
A1
Note: Accept .
[2 marks]
attempt to integrate between 0 and 30 (M1)
Note: An unsupported answer of 38.6 can imply integrating from 0 to 30.
EITHER
(A1)
OR
(A1)
THEN
A1
[3 marks]
Examiners report
Question
The following diagram shows part of the graph of for .
The shaded region is the area bounded by the curve, the -axis and the lines and .
Using implicit differentiation, find an expression for .
Find the equation of the tangent to the curve at the point .
Find the area of .
The region is now rotated about the -axis, through radians, to form a solid.
By writing as , show that the volume of the solid formed is .
Markscheme
valid attempt to differentiate implicitly (M1)
A1A1
A1
[4 marks]
at (M1)
A1
hence equation of tangent is
OR (M1)A1
Note: Accept .
[4 marks]
(M1)
(A1)
A1
[3 marks]
use of volume (M1)
A1
Note: Condone absence of limits up to this point.
reasonable attempt to integrate (M1)
A1A1
Note: Award A1 for correct limits (not to be awarded if previous M1 has not been awarded) and A1 for correct integrand.
A1
AG
Note: Do not accept decimal answer equivalent to .
[6 marks]
Examiners report
Question
Assuming the Maclaurin series for and , show that the Maclaurin series for is
By differentiating the series in part (a), show that the Maclaurin series for is .
Hence determine the Maclaurin series for as far as the term in .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
attempts to substitute into
M1
A1
attempts to expand the up to and including the term M1
A1
AG
METHOD 2
attempts to substitute into M1
attempts to find the Maclaurin series for up to and including the term M1
A1
A1
AG
[4 marks]
A1A1
attempts to expand the up to and including the term M1
A1
AG
[4 marks]
METHOD 1
let
uses to form M1
A1
(A1)
attempts to equate coefficients,
M1
A1
so
METHOD 2
uses to form M1
A1
(A1)
attempts to expand the up to and including the term M1
A1
Note: Accept use of long division.
[5 marks]
Examiners report
Question
Consider the differential equation , where .
It is given that when .
Solve the differential equation, giving your answer in the form .
The graph of against has a local maximum between and . Determine the coordinates of this local maximum.
Show that there are no points of inflexion on the graph of against .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
puts so that M1
A1
attempts to express as a single rational fraction in
M1
attempts to separate variables M1
A1A1
substitutes and attempts to find the value of M1
A1
the solution is
A1
[9 marks]
at a maximum, M1
attempts to substitute into their solution M1
attempts to solve for (M1)
A1
Note: Accept all answers that round to the correct answer.
Accept .
[4 marks]
METHOD 1
attempts (quotient rule) implicit differentiation M1
correctly substitutes into
A1
A1
this expression can never be zero therefore no points of inflexion R1
METHOD 2
attempts implicit differentiation on M1
A1
A1
and therefore no points of inflexion R1
Note: Accept putting and obtaining contradiction.
[4 marks]
Examiners report
Question
The following diagram shows part of the graph of . The graph has a local maximum point at and a local minimum point at .
Determine the values of , and .
Hence find the area of the shaded region.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
the principal axis is
so A1
the amplitude is
so A1
EITHER
one period is (M1)
OR
Substituting a point eg
Choice of correct solution (M1)
THEN
A1
Note: and can be both given as negatives for full marks
[4 marks]
roots are (A1)
(M1)
(A1)
so area A1
[4 marks]
Examiners report
Question
A small bead is free to move along a smooth wire in the shape of the curve .
Find an expression for .
At the point on the curve where , it is given that
Find the value of at this exact same instant.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid attempt to use chain rule or quotient rule (M1)
OR A1A1
[3 marks]
Note: Award A1 for numerator and A1 for denominator, or A1 for each part if the second alternative given.
valid attempt to use chain rule (M1)
or equivalent (A1)
A1
[3 marks]
Examiners report
Question
A particle moves in a straight line such that after time seconds, its velocity, in , is given by , where .
At time , has displacement ; at time , .
At successive times when the acceleration of is, the velocities of form a geometric sequence. The acceleration of is zero at times where and the respective velocities are .
Find the times when comes to instantaneous rest.
Find an expression for in terms of .
Find the maximum displacement of , in metres, from its initial position.
Find the total distance travelled by in the first seconds of its motion.
Show that, at these times, .
Hence show that .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
A1
A1
[2 marks]
attempt to use integration by parts M1
EITHER
A1
A1
M1
OR
A1
A1
M1
THEN
A1
at M1
A1
[7 marks]
EITHER
substituting into their equation for (M1)
OR
using GDC to find maximum value (M1)
OR
evaluating (M1)
THEN
A1
[2 marks]
METHOD 1
EITHER
distance required (M1)
OR
distance required (M1)
THEN
A1
METHOD 2
using successive minimum and maximum values on the displacement graph (M1)
A1
[2 marks]
valid attempt to find using product rule and set M1
A1
AG
[2 marks]
attempt to evaluate in exact form M1
A1
Note: The A1 is for any two consecutive correct, or showing that or .
showing that
eg M1A1
showing that M1
eg
Note: Award the A1 for any two consecutive terms.
AG
[5 marks]
Examiners report
Question
A continuous random variable has the probability density function given by
where .
Show that .
Find the value of .
Markscheme
recognition of the need to integrate (M1)
EITHER
(or equivalent) (A1)
A1
OR
(A1)
A1
THEN
attempt to use correct limits for their integrand and set equal to M1
OR
A1
AG
[5 marks]
attempt to solve (M1)
A1
[2 marks]
Examiners report
Question
Write down the first three terms of the binomial expansion of in ascending powers of .
By using the Maclaurin series for and the result from part (a), show that the Maclaurin series for up to and including the term in is .
By using the Maclaurin series for and the result from part (b), find .
Markscheme
A1
Note: Accept and .
[1 mark]
(M1)
or (M1)
A1
A1
so the Maclaurin series for up to and including the term in is AG
Note: Condone the absence of ‘…’
[4 marks]
M1
A1
A1
Note: Condone missing ‘lim’ and errors in higher derivatives.
Do not award M1 unless is replaced by in .
[3 marks]
Examiners report
Question
Two boats and travel due north.
Initially, boat is positioned metres due east of boat .
The distances travelled by boat and boat , after seconds, are metres and metres respectively. The angle is the radian measure of the bearing of boat from boat . This information is shown on the following diagram.
Show that .
At time , the following conditions are true.
Boat has travelled metres further than boat .
Boat is travelling at double the speed of boat .
The rate of change of the angle is radians per second.
Find the speed of boat at time .
Markscheme
OR A1
AG
Note: may be identified as a length on a diagram, and not written explicitly.
[1 mark]
attempt to differentiate with respect to (M1)
A1
attempt to set speed of equal to double the speed of (M1)
A1
OR (A1)
Note: This A1 can be awarded independently of previous marks.
So the speed of boat is A1
Note: Accept from the use of inexact values.
[6 marks]
Examiners report
Question
A function is defined by , where and .
The region enclosed by the graph of , the -axis, the -axis and the line is rotated about the -axis to form a solid of revolution.
Pedro wants to make a small bowl with a volume of based on the result from part (a). Pedro’s design is shown in the following diagrams.
The vertical height of the bowl, , is measured along the -axis. The radius of the bowl’s top is and the radius of the bowl’s base is . All lengths are measured in .
For design purposes, Pedro investigates how the cross-sectional radius of the bowl changes.
Show that the volume of the solid formed is cubic units.
Find the value of that satisfies the requirements of Pedro’s design.
Find .
Find .
By sketching the graph of a suitable derivative of , find where the cross-sectional radius of the bowl is decreasing most rapidly.
State the cross-sectional radius of the bowl at this point.
Markscheme
attempt to use (M1)
EITHER
applying integration by recognition (M1)
A3
OR
(A1)
attempt to express the integral in terms of (M1)
when and when
(A1)
A1
OR
(A1)
attempt to express the integral in terms of (M1)
when and when
(A1)
A1
Note: Accept equivalent working with indefinite integrals and original limits for .
THEN
A1
so the volume of the solid formed is cubic units AG
Note: Award (M1)(A0)(M0)(A0)(A0)(A1) when is obtained from GDC
[6 marks]
a valid algebraic or graphical attempt to find (M1)
(as ) A1
Note: Candidates may use their GDC numerical solve feature.
[2 marks]
attempting to find
with (M1)
A1
[2 marks]
attempting to find
with (M1)
A1
[2 marks]
EITHER
recognising to graph (M1)
Note: Award M1 for attempting to use quotient rule or product rule differentiation.
for graph decreasing to the local minimum A1
before increasing towards the -axis A1
OR
recognising to graph (M1)
Note: Award M1 for attempting to use quotient rule or product rule differentiation.
for , graph increasing towards and beyond the -intercept A1
recognising for maximum rate (A1)
THEN
A1
Note: Only award A marks if either graph is seen.
[4 marks]
attempting to find (M1)
the cross-sectional radius at this point is A1
[2 marks]
Examiners report
Question
The function is defined by , for , , .
The graph of has exactly one point of inflexion.
The function is defined by , for .
Find the value of and the value of .
Find an expression for .
Find the -coordinate of the point of inflexion.
Sketch the graph of for , showing the values of any axes intercepts, the coordinates of any local maxima and local minima, and giving the equations of any asymptotes.
Find the equations of all the asymptotes on the graph of .
By considering the graph of , or otherwise, solve for .
Markscheme
attempt to solve e.g. by factorising (M1)
or vice versa A1
[2 marks]
attempt to use quotient rule or product rule (M1)
EITHER
A1A1
Note: Award A1 for each term in the numerator with correct signs, provided correct denominator is seen.
OR
A1A1
Note: Award A1 for each term.
[3 marks]
attempt to find the local min point on OR solve (M1)
A1
[2 marks]
A1A1A1A1A1
Note: Award A1 for both vertical asymptotes with their equations, award A1 for horizontal asymptote with equation, award A1 for each correct branch including asymptotic behaviour, coordinates of minimum and maximum points (may be seen next to the graph) and values of axes intercepts.
If vertical asymptotes are absent (or not vertical) and the branches overlap as a consequence, award maximum A0A1A0A1A1.
[5 marks]
A1
(oblique asymptote has) gradient (A1)
appropriate method to find complete equation of oblique asymptote M1
A1
Note: Do not award the final A1 if the answer is not given as an equation.
[4 marks]
attempting to find at least one critical value (M1)
OR OR A1A1A1
Note: Only penalize once for use of rather than .
[4 marks]
Examiners report
Question
The function has a derivative given by where is a positive constant.
Consider , the population of a colony of ants, which has an initial value of .
The rate of change of the population can be modelled by the differential equation , where is the time measured in days, , and is the upper bound for the population.
At the population of the colony has doubled in size from its initial value.
The expression for can be written in the form , where . Find and in terms of .
Hence, find an expression for .
By solving the differential equation, show that .
Find the value of , giving your answer correct to four significant figures.
Find the value of when the rate of change of the population is at its maximum.
Markscheme
(A1)
attempt to compare coefficients OR substitute and and solve (M1)
and A1
[3 marks]
attempt to integrate their (M1)
A1A1
Note: Award A1 for each correct term. Award A1A0 for a correct answer without modulus signs. Condone the absence of .
[3 marks]
attempt to separate variables and integrate both sides M1
A1
Note: There are variations on this which should be accepted, such as . Subsequent marks for these variations should be awarded as appropriate.
EITHER
attempt to substitute into an equation involving M1
A1
A1
A1
OR
A1
attempt to substitute M1
A1
A1
THEN
attempt to rearrange and isolate M1
OR OR
OR A1
AG
[8 marks]
attempt to substitute (M1)
(A1)
A1
Note: Award (M1)(A1)A0 for any other value of which rounds to
[3 marks]
attempt to find the maximum of the first derivative graph OR zero of the second derivative graph OR that (M1)
(days) A2
Note: Accept any value which rounds to .
[3 marks]
Examiners report
Question
A function is defined by .
A function is defined by .
Show that is an even function.
By considering limits, show that the graph of has a horizontal asymptote and state its equation.
Show that for .
By using the expression for and the result , show that is decreasing for .
Find an expression for , justifying your answer.
State the domain of .
Sketch the graph of , clearly indicating any asymptotes with their equations and stating the values of any axes intercepts.
Markscheme
EITHER
R1
OR
a sketch graph of with line symmetry in the -axis indicated R1
THEN
so is an even function. AG
[1 mark]
as A1
so the horizontal asymptote is A1
[2 marks]
attempting to use the quotient rule to find M1
A1
attempting to use the chain rule to find M1
let and so and
M1
A1
A1
AG
[6 marks]
EITHER
for (A1)
so A1
OR
and A1
A1
THEN
R1
Note: Award R1 for stating that in , the numerator is negative, and the denominator is positive.
so is decreasing for AG
Note: Do not accept a graphical solution
[3 marks]
M1
A1
A1
domain of is and so the range of must be
hence the positive root is taken (or the negative root is rejected) R1
Note: The R1 is dependent on the above A1.
so A1
Note: The final A1 is not dependent on R1 mark.
[5 marks]
domain is A1
Note: Accept correct alternative notations, for example, or .
Accept if correct to s.f.
[1 mark]
A1A1A1
Note: A1 for correct domain and correct range and -intercept at
A1 for asymptotic behaviour
A1 for
Coordinates are not required.
Do not accept or other inexact values.
[3 marks]
Examiners report
Question
Consider the curve given by where .
Show that .
Hence find the equation of the tangent to at the point where .
Markscheme
METHOD 1
attempts to differentiate implicitly including at least one application of the product rule (M1)
A1
Note: Award (M1)A1 for implicitly differentiating and obtaining .
A1
AG
METHOD 2
attempts to differentiate implicitly including at least one application of the product rule (M1)
A1
or equivalent to the above, for example
A1
or equivalent to the above, for example
AG
METHOD 3
attempt to differentiate implicitly including at least one application of the product rule M1
A1
A1
AG
METHOD 4
lets and attempts to find where M1
A1
A1
AG
[3 marks]
METHOD 1
substitutes into (M1)
A1
substitutes and their non-zero value of into (M1)
A1
equation of the tangent is A1
METHOD 2
substitutes into (M1)
EITHER
correctly substitutes into A1
A1
OR
correctly substitutes into A1
A1
THEN
substitutes into (M1)
equation of the tangent is A1
[5 marks]
Examiners report
Question
Consider the function .
Find the coordinates where the graph of crosses the
-axis.
-axis.
Write down the equation of the vertical asymptote of the graph of .
The oblique asymptote of the graph of can be written as where .
Find the value of and the value of .
Sketch the graph of for , clearly indicating the points of intersection with each axis and any asymptotes.
Express in partial fractions.
Hence find the exact value of , expressing your answer as a single logarithm.
Markscheme
Note: In part (a), penalise once only, if correct values are given instead of correct coordinates.
attempts to solve (M1)
and A1
[2 marks]
Note: In part (a), penalise once only, if correct values are given instead of correct coordinates.
A1
[1 mark]
A1
Note: Award A0 for .
Award A1 in part (b), if is seen on their graph in part (d).
[1 mark]
METHOD 1
attempts to expand (M1)
A1
equates coefficients of (M1)
A1
METHOD 2
attempts division on M1
M1
A1
A1
METHOD 3
A1
M1
equates coefficients of : (M1)
A1
METHOD 4
attempts division on M1
A1
M1
A1
[4 marks]
two branches with approximately correct shape (for ) A1
their vertical and oblique asymptotes in approximately correct positions with both branches showing correct asymptotic behaviour to these asymptotes A1
their axes intercepts in approximately the correct positions A1
Note: Points of intersection with the axes and the equations of asymptotes are not required to be labelled.
[3 marks]
attempts to split into partial fractions: (M1)
A1
A1
[3 marks]
attempts to integrate and obtains two terms involving ‘ln’ (M1)
A1
A1
A1
Note: The final A1 is dependent on the previous two A marks.
[4 marks]
Examiners report
Question
The following diagram shows the curve , where .
The curve from point to point is rotated about the -axis to form the interior surface of a bowl. The rectangle , of height , is rotated about the -axis to form a solid base.
The bowl is assumed to have negligible thickness.
Given that the interior volume of the bowl is to be , determine the height of the base.
Markscheme
attempts to express in terms of (M1)
A1
Note: Correct limits are required.
Attempts to solve for (M1)
Note: Award M1 for attempting to solve or equivalent for .
(cm) A2
[5 marks]
Examiners report
This question was a struggle for many candidates. To start with, many candidates had difficulty understanding the diagram. Some candidates tried to include the base in their equation.
Because of this confusion, the question was poorly attempted. Some only received one mark for rearranging the equation to make the subject but were unable to set the correct definite integral with correct terminals. Again, many candidates tried to solve by hand instead of using their GDC. The correct answer was not seen that often.
Those candidates who recognised that the volume was around the y -axis and used their GDC to solve, usually achieved full marks for this question.
Question
Consider , where .
Show that a finite limit only exists for .
Using l’Hôpital’s rule, show algebraically that the value of the limit is .
Markscheme
(as , the indeterminate form is required for the limit to exist)
M1
A1
so AG
Note: Award M1A0 for using to show the limit is .
[2 marks]
A1A1
Note: Award A1 for a correct numerator and A1 for a correct denominator.
recognises to apply l’Hôpital’s rule again (M1)
Note: Award M0 if their limit is not the indeterminate form .
EITHER
A1A1
Note: Award A1 for a correct first term in the numerator and A1 for a correct second term in the numerator.
OR
A1A1
Note: Award A1 for a correct numerator and A1 for a correct denominator.
THEN
substitutes into the correct expression to evaluate the limit A1
Note: The final A1 is dependent on all previous marks.
AG
[6 marks]
Examiners report
Part (a) Many candidates recognised the indeterminate form and provided a nice algebraic proof. Some verified by substituting the given value. Therefore, there is a need to teach the candidates the difference between proof and verification. Only a few candidates were able to give a complete 'show that' proof.
Part (b) Many candidates realised that they needed to apply the L'Hôpital's rule twice. There were many mistakes in differentiation using the chain rule. Not all candidates clearly showed the final substitution.
Question
A scientist conducted a nine-week experiment on two plants, and , of the same species. He wanted to determine the effect of using a new plant fertilizer. Plant was given fertilizer regularly, while Plant was not.
The scientist found that the height of Plant , at time weeks can be modelled by the function , where .
The scientist found that the height of Plant , at time weeks can be modelled by the function , where .
Use the scientist’s models to find the initial height of
Plant .
Plant correct to three significant figures.
Find the values of when .
For , prove that Plant was always taller than Plant .
For , find the total amount of time when the rate of growth of Plant was greater than the rate of growth of Plant .
Markscheme
(cm) A1
[1 mark]
(M1)
(cm) A1
[2 marks]
attempts to solve for (M1)
(weeks) A2
[3 marks]
A1
EITHER
for A1
and as R1
OR
the minimum value of R1
so for A1
THEN
hence for , Plant was always taller than Plant AG
[3 marks]
recognises that and are required (M1)
attempts to solve for (M1)
and OR and OR and (A1)
Note: Award full marks for .
Award subsequent marks for correct use of these exact values.
OR OR (A1)
attempts to calculate the total amount of time (M1)
(weeks) A1
[6 marks]
Examiners report
Part (a) In general, very well done, most students scored full marks. Some though had an incorrect answer for part(a)(ii) because they had their GDC in degrees.
Part (b) Well attempted. Some accuracy errors and not all candidates listed all three values.
Part (c) Most students tried a graphical approach (but this would only get them one out of three marks) and only some provided a convincing algebraic justification. Many candidates tried to explain in words without a convincing mathematical justification or used numerical calculations with specific time values. Some arrived at the correct simplified equation for the difference in heights but could not do much with it. Then only a few provided a correct mathematical proof.
Part (d) In general, well attempted by many candidates. The common error was giving the answer as 3.15 due to the pre-mature rounding. Some candidates only provided the values of time when the rates are equal, some intervals rather than the total time.
Question
Consider the function , where .
The curve is rotated about the -axis to form a solid of revolution that is used to model a water container.
At , the container is empty. Water is then added to the container at a constant rate of .
Sketch the curve , clearly indicating the coordinates of the endpoints.
Show that the inverse function of is given by .
State the domain and range of .
Show that the volume, , of water in the container when it is filled to a height of metres is given by .
Hence, determine the maximum volume of the container.
Find the time it takes to fill the container to its maximum volume.
Find the rate of change of the height of the water when the container is filled to half its maximum volume.
Markscheme
correct shape (concave down) within the given domain A1
and A1
Note: The coordinates of endpoints may be seen on the graph or marked on the axes.
[2 marks]
interchanging and (seen anywhere) M1
A1
A1
AG
[3 marks]
OR domain A1
OR OR range A1
[2 marks]
attempt to substitute into the correct volume formula (M1)
A1
A1
AG
Note: Award marks as appropriate for correct work using a different variable e.g.
[3 marks]
attempt to substitute into (M1)
A1
[2 marks]
time (M1)
A1
[2 marks]
attempt to find the height of the tank when (M1)
(A1)
attempt to use the chain rule or differentiate with respect to (M1)
OR (A1)
attempt to substitute their and (M1)
A1
[6 marks]
Examiners report
Part a) was generally well done, with the most common errors being to use an incorrect domain or not to give the coordinates of the endpoints. Some graphs appeared to be straight lines; some candidates drew sketches which were too small which made it more difficult for them to show the curvature.
Most candidates were able to show the steps to find an inverse function in part b), although occasionally a candidate did not explicitly swop the x and y variables before writing down the inverse function, which was given in the question. Many candidates struggled to identify the domain and range of the inverse, despite having a correct graph.
Part c) required a rotation around the y-axis, but a number of candidates attempted to rotate around the x-axis or failed to include limits. In the same vein, many substituted 2 into the formula instead of the square root of 3 when answering the second part. Many subsequently gained follow through marks on part d).
There were a number of good attempts at related rates in part e), with the majority differentiating V with respect to t, using implicit differentiation. However, many did not find the value of h which corresponded to halving the volume, and a number did not differentiate with respect to t, only with respect to h.
Question
Two airplanes, and , have position vectors with respect to an origin given respectively by
where represents the time in minutes and .
Entries in each column vector give the displacement east of , the displacement north of and the distance above sea level, all measured in kilometres.
The two airplanes’ lines of flight cross at point .
Find the three-figure bearing on which airplane is travelling.
Show that airplane travels at a greater speed than airplane .
Find the acute angle between the two airplanes’ lines of flight. Give your answer in degrees.
Find the coordinates of .
Determine the length of time between the first airplane arriving at and the second airplane arriving at .
Let represent the distance between airplane and airplane for .
Find the minimum value of .
Markscheme
let be the required angle (bearing)
EITHER
(M1)
Note: Award M1 for a labelled sketch.
OR
(M1)
THEN
A1
Note: Do not accept or or .
[2 marks]
METHOD 1
let be the speed of and let be the speed of
attempts to find the speed of one of or (M1)
or
Note: Award M0 for and .
(km min-1) and (km min-1) A1
so travels at a greater speed than AG
METHOD 2
attempts to use
and (M1)
for example:
and
and
and A1
so travels at a greater speed than AG
[2 marks]
attempts to use the angle between two direction vectors formula (M1)
(A1)
or
attempts to find the acute angle using their value of (M1)
A1
[4 marks]
for example, sets and forms at least two equations (M1)
Note: Award M0 for equations involving only.
EITHER
attempts to solve the system of equations for one of or (M1)
or A1
OR
attempts to solve the system of equations for and (M1)
or A1
THEN
substitutes their or value into the corresponding or (M1)
A1
Note: Accept . Accept km east of , km north of and km above sea level.
[5 marks]
attempts to find the value of (M1)
minutes ( seconds) A1
[2 marks]
EITHER
attempts to find (M1)
attempts to find their (M1)
A1
OR
attempts to find (M1)
attempts to find their (M1)
A1
Note: Award M0M0A0 for expressions using two different time parameters.
THEN
either attempts to find the local minimum point of or attempts to find the value of such that (or equivalent) (M1)
minimum value of is (km) A1
Note: Award M0 for attempts at the shortest distance between two lines.
[5 marks]
Examiners report
General comment about this question: many candidates were not exposed to this setting of vectors question and were rather lost.
Part (a) Probably the least answered question on the whole paper. Many candidates left it blank, others tried using 3D vectors. Out of those who calculated the angle correctly, only a small percentage were able to provide the correct true bearing as a 3-digit figure.
Part (b) Well done by many candidates who used the direction vectors to calculate and compare the speeds. A number of candidates tried to use the average rate of change but mostly unsuccessfully.
Part (c) Most candidates used the correct vectors and the formula to obtain the obtuse angle. Then only some read the question properly to give the acute angle in degrees, as requested.
Part (d) Well done by many candidates who used two different parameters. They were able to solve and obtain two values for time, the difference in minutes and the correct point of intersection. A number of candidates only had one parameter, thus scoring no marks in part (d) (i). The frequent error in part (d)(ii) was providing incorrect units.
Part (e) Many correct answers were seen with an efficient way of setting the question and using their GDC to obtain the answer, graphically or numerically. Some gave time only instead of actually giving the minimal distance. A number of candidates tried to find the distance between two skew lines ignoring the fact that the lines intersect.
Question
Consider the differential equation for and . It is given that when .
Use Euler’s method, with a step length of , to find an approximate value of when .
Use the substitution to show that .
By solving the differential equation, show that .
Find the actual value of when .
Using the graph of , suggest a reason why the approximation given by Euler’s method in part (a) is not a good estimate to the actual value of at .
Markscheme
attempt to use Euler’s method (M1)
, where
correct intermediate -values (A1)(A1)
Note: A1 for any two correct -values seen
A1
Note: For the final A1, the value must be the last value in a table or a list, or be given as a final answer, not just embedded in a table which has further lines.
[4 marks]
(A1)
replacing with and with M1
A1
(since )
AG
[3 marks]
attempt to separate variables and (M1)
(A1)
attempt to express in partial fraction form M1
A1
A1
Note: Condone absence of modulus signs throughout.
EITHER
attempt to find using M1
expressing both sides as a single logarithm (M1)
OR
expressing both sides as a single logarithm (M1)
attempt to find using M1
THEN
(since )
substitute (seen anywhere) M1
(since )
attempt to make the subject M1
A1
AG
[10 marks]
actual value at A1
[1 mark]
gradient changes rapidly (during the interval considered) OR
the curve has a vertical asymptote at R1
[1 mark]
Examiners report
Most candidates showed evidence of an attempt to use Euler's method in part a), although very few explicitly wrote down the formulae, they used in order to calculate successive y-value. In addition, many seemed to take a step-by-step approach rather than using the recursive capabilities of the graphical display calculator.
There were many good attempts at part b), but not all candidates recognised that this would help them to solve part c).
Part c) was done very well by many candidates, although there were a significant number who failed to recognise the need for partial fractions and could not make further progress. A common error was to integrate without a constant of integration, which meant that the initial condition could not be used. The reasoning given for the estimate being poor was often too vague and did not address the specific nature of the function given clearly enough.
Question
The population, , of a particular species of marsupial on a small remote island can be modelled by the logistic differential equation
where is the time measured in years and are positive constants.
The constant represents the maximum population of this species of marsupial that the island can sustain indefinitely.
Let be the initial population of marsupials.
In the context of the population model, interpret the meaning of .
Show that .
Hence show that the population of marsupials will increase at its maximum rate when . Justify your answer.
Hence determine the maximum value of in terms of and .
By solving the logistic differential equation, show that its solution can be expressed in the form
.
After years, the population of marsupials is . It is known that .
Find the value of for this population model.
Markscheme
rate of growth (change) of the (marsupial) population (with respect to time) A1
[1 mark]
Note: Do not accept growth (change) in the (marsupials) population per year.
METHOD 1
attempts implicit differentiation on be expanding (M1)
A1A1
A1
and so AG
METHOD 2
attempts implicit differentiation (product rule) on M1
A1
substitutes into their M1
A1
so AG
[4 marks]
(M1)
A2
Note: Award A1 for only.
uses the second derivative to show that concavity changes at or the first derivative to show a local maximum at M1
EITHER
a clearly labelled correct sketch of versus showing corresponding to a local maximum point for R1
OR
a correct and clearly labelled sign diagram (table) showing corresponding to a local maximum point for R1
OR
for example, with and with showing corresponds to a local maximum point for R1
so the population is increasing at its maximum rate when AG
[5 marks]
substitutes into (M1)
the maximum value of is A1
[2 marks]
METHOD 1
attempts to separate variables M1
attempts to write in partial fractions form M1
A1
A1A1
Note: Award A1 for and A1 for and . Absolute value signs are not required.
attempts to find in terms of and M1
when and so
A1
so AG
METHOD 2
attempts to separate variables M1
attempts to write in partial fractions form M1
A1
A1A1
Note: Award A1 for and A1 for and . Absolute value signs are not required.
attempts to find in terms of and M1
when and so
A1
AG
METHOD 3
lets and forms M1
multiplies both sides of the differential equation by and makes the above substitutions M1
(linear first-order DE) A1
(M1)
A1
attempts to find in terms of and M1
when and so
A1
AG
[7 marks]
substitutes and into M1
A1
[2 marks]
Examiners report
An extremely tricky question even for the strong candidates. Many struggled to understand what was expected in parts (b) and (c). As the question was set all with pronumerals instead of numbers many candidates found it challenging, thrown at deep water for parts (b), (c) and (e). It definitely was the question to show their skills for the Level 7 candidates provided that they did not run out of time.
Part (a) Very well answered, mostly correctly referring to the rate of change. Some candidates did not gain this mark because their sentence did not include the reference to the rate of change. Worded explanations continue being problematic to many candidates.
Part (b) Many candidates were confused how to approach this question and did not realise that they
needed to differentiate implicitly. Some tried but with errors, some did not fully show what was required.
Part (c) Most candidates started with equating the second derivative to zero. Most gave the answer omitting the other two possibilities. Most stopped here. Only a small number of candidates provided the correct mathematical argument to show it is a local maximum.
Part (d) Well done by those candidates who got that far. Most got the correct answer, sometimes not fully simplified.
Part (e) Most candidates separated the variables, but some were not able to do much more. Some candidates knew to resolve into partial fractions and attempted to do so, mainly successfully. Then they integrated, again, mainly successfully and continued to substitute the initial condition and manipulate the equation accordingly.
Part (f) Algebraic manipulation of the logarithmic expression was too much for some candidates with a common error of 0.33 given as the answer. The strong candidates provided the correct exact or rounded value.
Question
The curve has equation .
Show that .
The tangent to at the point Ρ is parallel to the -axis.
Find the -coordinate of Ρ.
Markscheme
* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
attempts implicit differentiation on both sides of the equation M1
A1
A1
so AG
[3 marks]
attempts to solve for (M1)
A1
attempts to solve for given their value of (M1)
A1
[4 marks]
Examiners report
Question
Consider the differential equation
The curve for has a gradient function given by
.
The curve passes through the point .
Use the substitution to show that where is an arbitrary constant.
By using the result from part (a) or otherwise, solve the differential equation and hence show that the curve has equation .
The curve has a point of inflexion at where . Determine the coordinates of this point of inflexion.
Use the differential equation to show that the points of zero gradient on the curve lie on two straight lines of the form where the values of are to be determined.
Markscheme
* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
M1
A1
A1
integrating the RHS, AG
[3 marks]
EITHER
attempts to find M1
(A1)
substitutes their into M1
attempts to complete the square (M1)
A1
A1
OR
attempts to find M1
A1
M1
attempts to complete the square (M1)
A1
A1
THEN
when , (or ) and so M1
substitutes for into their expression M1
A1
so AG
[9 marks]
METHOD 1
EITHER
a correct graph of (for approximately ) with a local minimum point below the -axis A2
Note: Award M1A1 for .
attempts to find the -coordinate of the local minimum point on the graph of (M1)
OR
a correct graph of (for approximately ) showing the location of the -intercept A2
Note: Award M1A1 for .
attempts to find the -intercept (M1)
THEN
A1
attempts to find (M1)
the coordinates are A1
METHOD 2
attempts implicit differentiation on to find M1
(or equivalent)
() A1
attempts to solve for where M1
A1
attempts to find (M1)
the coordinates are A1
[6 marks]
M1
attempts to solve for M1
or A1
and A1
Note: Award M1 for stating , M1 for substituting into , A1 for and A1 for and .
[4 marks]
Examiners report
Question
A large tank initially contains pure water. Water containing salt begins to flow into the tank The solution is kept uniform by stirring and leaves the tank through an outlet at its base. Let grams represent the amount of salt in the tank and let minutes represent the time since the salt water began flowing into the tank.
The rate of change of the amount of salt in the tank, , is described by the differential equation .
Show that + 1 is an integrating factor for this differential equation.
Hence, by solving this differential equation, show that .
Sketch the graph of versus for 0 ≤ ≤ 60 and hence find the maximum amount of salt in the tank and the value of at which this occurs.
Find the value of at which the amount of salt in the tank is decreasing most rapidly.
The rate of change of the amount of salt leaving the tank is equal to .
Find the amount of salt that left the tank during the first 60 minutes.
Markscheme
METHOD 1
M1
= A1
AG
METHOD 2
attempting product rule differentiation on M1
A1
so is an integrating factor for this differential equation AG
[2 marks]
attempting to multiply through by and rearrange to give (M1)
A1
A1
attempting to integrate the RHS by parts M1
A1
Note: Condone the absence of C.
EITHER
substituting M1
A1
using as the highest common factor of and M1
OR
using as the highest common factor of and giving
(or equivalent) M1A1
substituting M1
THEN
AG
[8 marks]
graph starts at the origin and has a local maximum (coordinates not required) A1
sketched for 0 ≤ ≤ 60 A1
correct concavity for 0 ≤ ≤ 60 A1
maximum amount of salt is 14.6 (grams) at = 6.60 (minutes) A1A1
[5 marks]
using an appropriate graph or equation (first or second derivative) M1
amount of salt is decreasing most rapidly at = 12.9 (minutes) A1
[2 marks]
EITHER
attempting to form an integral representing the amount of salt that left the tank M1
A1
OR
attempting to form an integral representing the amount of salt that entered the tank minus the amount of salt in the tank at = 60(minutes)
amount of salt that left the tank is A1
THEN
= 36.7 (grams) A2
[4 marks]
Examiners report
Question
Consider the differential equation , given that when .
Show that is an integrating factor for this differential equation.
Hence solve this differential equation. Give the answer in the form .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
attempting to find an integrating factor (M1)
(M1)A1
IF is (M1)A1
AG
METHOD 2
multiply by the integrating factor
M1A1
left hand side is equal to the derivative of
A3
[5 marks]
(M1)
A1A1
M1A1
A1
[6 marks]
Examiners report
Question
Use l’Hôpital’s rule to determine the value of
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to use l’Hôpital’s rule, M1
or A1A1
Note: Award A1 for numerator A1 for denominator.
this gives 0/0 so use the rule again (M1)
or A1A1
Note: Award A1 for numerator A1 for denominator.
A1
Note: This A1 is dependent on all previous marks being awarded, except when the first application of L’Hopital’s does not lead to 0/0, when it should be awarded for the correct limit of their derived function.
[7 marks]
Examiners report
Question
Consider the differential equation
Use the substitution to show that the general solution of this differential equation is
Hence, or otherwise, solve the differential equation
given that when . Give your answer in the form .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
M1
the differential equation becomes
A1
A1
integrating, Constant AG
[3 marks]
EITHER
(A1)
M1A1
A1
Note: A1 is for correct factorization.
A1
OR
A1
M1
(A1)
Note: A1 is for correct factorization.
A1A1
THEN
substitute or when (M1)
therefore A1
Note: This A1 can be awarded anywhere in their solution.
substituting for ,
M1
Note: Award for correct substitution of into their expression.
(A1)
Note: Award for any rearrangement of a correct expression that has in the numerator.
A1
[10 marks]
Examiners report
Question
Consider the differential equation where when .
Show that is an integrating factor for this differential equation.
Solve the differential equation giving your answer in the form .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
integrating factor (M1)
(M1)
Note: Award M1 for use of for example or .
integrating factor A1
A1
Note: Award A1 for where .
AG
METHOD 2
M1A1
M1A1
Note: Award M1 for attempting to express in the form .
so is an integrating factor for this differential equation AG
[4 marks]
(or equivalent) (M1)
A1
(M1)A1
Note: Award M1 for using an appropriate substitution.
Note: Condone the absence of .
substituting M1
Note: Award M1 for attempting to find their value of .
A1
[6 marks]
Examiners report
Question
Find the value of .
Illustrate graphically the inequality .
Hence write down a lower bound for .
Find an upper bound for .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)
Note: The above A1 for using a limit can be awarded at any stage.
Condone the use of .
Do not award this mark to candidates who use as the upper limit throughout.
= M1
A1
[3 marks]
A1A1A1A1
A1 for the curve
A1 for rectangles starting at
A1 for at least three upper rectangles
A1 for at least three lower rectangles
Note: Award A0A1 for two upper rectangles and two lower rectangles.
sum of areas of the lower rectangles < the area under the curve < the sum of the areas of the upper rectangles so
AG
[4 marks]
a lower bound is A1
Note: Allow FT from part (a).
[1 mark]
METHOD 1
(M1)
(M1)
, an upper bound A1
Note: Allow FT from part (a).
METHOD 2
changing the lower limit in the inequality in part (b) gives
(A1)
(M1)
, an upper bound A1
Note: Condone candidates who do not use a limit.
[3 marks]
Examiners report
Question
The function is defined by .
The function satisfies the equation .
Show that .
By differentiating the above equation twice, show that
where and denote the 3rd and 4th derivative of respectively.
Hence show that the Maclaurin series for up to and including the term in is .
Use this series approximation for with to find an approximate value for .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
M1A1
Note: Award M1 for an attempt at chain rule differentiation.
Award M0A0 for .
AG
[2 marks]
differentiating gives M1A1
differentiating again gives M1A1
Note: Award M1 for an attempt at product rule differentiation of at least one product in each of the above two lines.
Do not penalise candidates who use poor notation.
AG
[4 marks]
attempting to find one of , or by substituting into relevant differential equation(s) (M1)
Note: Condone found by calculating at .
and A1
and so A1
Note: Only award the above A1, for correct first differentiation in part (b) leading to stated or seen from use of the general Maclaurin series.
Special Case: Award (M1)A0A1 if is stated without justification or found by working backwards from the general Maclaurin series.
so the Maclaurin series for up to and including the term in is AG
[3 marks]
substituting into M1
the series approximation gives a value of
so
A1
Note: Accept 9.76.
[2 marks]
Examiners report
Question
Consider the differential equation where and is a positive integer, .
Solve the differential equation given that when . Give your answer in the form .
Show that the -coordinate(s) of the points on the curve where satisfy the equation .
Deduce the set of values for such that there are two points on the curve where . Give a reason for your answer.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
(M1)
integrating factor M1
(A1)
= A1
(M1)
A1
Note: Condone the absence of C.
substituting , M1
Note: Award M1 for attempting to find their value of C.
A1
[8 marks]
METHOD 2
put so that M1(A1)
substituting, M1
(A1)
M1
A1
Note: Condone the absence of C.
substituting , M1
Note: Award M1 for attempting to find their value of C.
A1
[8 marks]
METHOD 1
find and solve for
M1
A1
Note: Award a maximum of M1A0 if a candidate’s answer to part (a) is incorrect.
AG
METHOD 2
substitute and their into the differential equation and solve for
M1
A1
Note: Award a maximum of M1A0 if a candidate’s answer to part (a) is incorrect.
AG
[2 marks]
there are two solutions for when is odd (and A1
if is even there are two solutions (to )
and if is odd there is only one solution (to ) R1
Note: Only award the R1 if both cases are considered.
[4 marks]
Examiners report
Question
Use L’Hôpital’s rule to determine the value of
Hence find .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
M1A1A1
A1
= −3 A1
[5 marks]
is of the form
applying l’Hôpital´s rule (M1)
(A1)
= −3 A1
[3 marks]
Examiners report
Question
Consider the differential equation , where .
Consider the family of curves which satisfy the differential equation , where .
Given that , use Euler’s method with step length = 0.25 to find an approximation for . Give your answer to two significant figures.
Solve the equation for .
Find the percentage error when is approximated by the final rounded value found in part (a). Give your answer to two significant figures.
Find the equation of the isocline corresponding to , where , .
Show that such an isocline can never be a normal to any of the family of curves that satisfy the differential equation.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to apply Euler’s method (M1)
(A1)(A1)
Note: Award A1 for correct values, A1 for first three correct values.
= 3.3 A1
[4 marks]
METHOD 1
(M1)
(A1)
(M1)
A1
M1
A1
METHOD 2
M1
(A1)
M1
A1
M1
A1
[6 marks]
percentage error (M1)(A1)
= 2.5% A1
[3 marks]
A1
[1 mark]
gradient of isocline equals gradient of normal (M1)
or A1
A1
R1
no solution AG
Note: Accept alternative reasons for no solutions.
[4 marks]
Examiners report
Question
A simple model to predict the population of the world is set up as follows. At time years the population of the world is , which can be assumed to be a continuous variable. The rate of increase of due to births is 0.056 and the rate of decrease of due to deaths is 0.035.
Show that .
Find a prediction for the number of years it will take for the population of the world to double.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
A1
AG
[1 mark]
METHOD 1
attempt to separate variables M1
A1
A1
EITHER
A1
Note: This A1 is independent of the following marks.
OR
A1
Note: This A1 is independent of the following marks.
THEN
(M1)
years A1
Note: If a candidate writes , so then award the final A1.
METHOD 2
attempt to separate variables M1
A1A1
Note: Award A1 for correct integrals and A1 for correct limits seen anywhere. Do not penalize use of in place of .
A1
(M1)
A1
[6 marks]
Examiners report
Question
Using L’Hôpital’s rule, find .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
M1A1A1
Note: Award M1 for attempt at differentiation using l'Hopital's rule, A1 for numerator, A1 for denominator.
METHOD 1
using l’Hopital’s rule again
A1A1
EITHER
A1A1
Note: Not all terms in numerator need to be written in final fraction. Award A1 for . However, if the terms are written, they
must be correct to award A1.
attempt to substitute M1
OR
(M1)A1
A1
THEN
A1
METHOD 2
M1
A1
M1A1
attempt to substitute M1
A1
[9 marks]
Examiners report
Question
Consider the differential equation , where .
Solve the differential equation and show that a general solution is where is a positive constant.
Prove that there are two horizontal tangents to the general solution curve and state their equations, in terms of .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
let M1
(A1)
(M1)
(A1)
Note: Or equivalent attempt at simplification.
A1
(M1)
(A1)
A1A1
Note: Award A1 for LHS and A1 for RHS and a constant.
M1
Note: Award M1 for substituting . May be seen at a later stage.
A1
Note: Award A1 for any correct equivalent equation without logarithms.
AG
[11 marks]
METHOD 1
(for horizontal tangents) M1
EITHER
using M1
A1
Note: Award M1A1 for .
OR
using implicit differentiation of
M1
Note: Accept differentiation of .
A1
THEN
tangents at A1A1
hence there are two tangents AG
METHOD 2
M1A1
this is a circle radius centre A1
hence there are two tangents AG
tangents at A1A1
[5 marks]
Examiners report
Question
The function is defined by , where .
By finding a suitable number of derivatives of , find the first two non-zero terms in the Maclaurin series for .
Hence or otherwise, find .
Markscheme
M1A1
Note: Award M1A0 for
A1
EITHER
A1
OR
A1
THEN
substitute into or any of its derivatives (M1)
, and A1
the Maclaurin series is
(M1)A1
[8 marks]
METHOD 1
M1
(M1)
A1
Note: Condone the omission of +… in their working.
METHOD 2
indeterminate form, using L’Hôpital’s rule
M1
indeterminate form, using L’Hôpital’s rule again
M1
Note: Award M1 only if their previous expression is in indeterminate form.
A1
Note: Award FT for use of their derivatives from part (a).
[3 marks]
Examiners report
Question
Consider the differential equation , with when .
Use Euler’s method, with step length , to find an approximate value of when .
Sketch the isoclines for .
Express in the form , where .
Solve the differential equation, for , giving your answer in the form .
Sketch the graph of for .
With reference to the curvature of your sketch in part (c)(iii), and without further calculation, explain whether you conjecture will be less than, equal to, or greater than your answer in part (a).
Markscheme
(M1)(A1)(A1)(A1)A1
Note: Award A1 for each correct value.
For the intermediate values, accept answers that are accurate to 2 significant figures.
The final value must be accurate to 3 significant figures or better.
[5 marks]
attempt to solve (M1)
or
A1A1
[3 marks]
A1
[1 mark]
recognition of homogeneous equation,
let M1
the equation can be written as
(A1)
M1
Note: Award M1 for attempt to separate the variables.
from part (c)(i) M1
A1A1
M1
Note: Award M1 for using initial conditions to find .
A1
substituting M1
Note: This M1 may be awarded earlier.
A1
[10 marks]
curve drawn over correct domain A1
[1 mark]
the sketch shows that is concave up A1
Note: Accept is increasing.
this means the tangent drawn using Euler’s method will give an underestimate of the real value, so > estimate in part (a) R1
Note: The R1 is dependent on the A1.
[2 marks]
Examiners report
Question
Use l’Hôpital’s rule to find
.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to use l’Hôpital’s rule M1
A1A1
Note: Award A1 for the numerator and A1 for the denominator.
substitution of into their expression (M1)
hence use l’Hôpital’s rule again
Note: If the first use of l’Hôpital’s rule results in an expression which is not in indeterminate form, do not award any further marks.
attempt to use product rule in numerator M1
A1
A1
[7 marks]
Examiners report
Question
The curve has a gradient function given by
.
The curve passes through the point .
On the same set of axes, sketch and label isoclines for and , and clearly indicate the value of each -intercept.
Hence or otherwise, explain why the point is a local minimum.
Find the solution of the differential equation , which passes through the point . Give your answer in the form .
Explain why the graph of does not intersect the isocline .
Sketch the graph of on the same set of axes as part (a)(i).
Markscheme
attempt to find equation of isoclines by setting M1
parallel lines with positive gradient A1
-intercept for A1
Note: To award A1, each -intercept should be clear, but condone a missing label (eg. ).
If candidates represent the lines using slope fields, but omit the lines, award maximum of M1A0A1.
[3 marks]
at point A1
EITHER
to the left of , the gradient is negative R1
to the right of , the gradient is positive R1
Note: Accept any correct reasoning using gradient, isoclines or slope field.
If a candidate uses left/right or without explicitly referring to the point or a correct region on the diagram, award R0R1.
OR
A1
A1
Note: accept correct reasoning that is increasing as increases.
THEN
hence is a local minimum AG
[3 marks]
integrating factor (M1)
(A1)
(M1)
A1
(M1)
A1
Note: Award A1 for the correct RHS.
substituting gives
M1
A1
[8 marks]
METHOD 1
EITHER
attempt to solve for the intersection (M1)
OR
attempt to find the difference (M1)
THEN
for all R1
Note: Accept or equivalent reasoning.
therefore the curve does not intersect the isocline AG
METHOD 2
is an (oblique) asymptote to the curve R1
Note: Do not accept “the curve is parallel to "
is the isocline for R1
therefore the curve does not intersect the isocline AG
METHOD 3
The initial point is above , so R1
R1
therefore the curve does not intersect the isocline AG
[2 marks]
concave up curve with minimum at approximately A1
asymptote of curve is isocline A1
Note: Only award FT from (b) if the above conditions are satisfied.
[2 marks]
Examiners report
Question
The function is defined by where .
The seventh derivative of is given by .
Use the Maclaurin series for to write down the first three non-zero terms of the Maclaurin series for .
Hence find the first three non-zero terms of the Maclaurin series for .
Use your answer to part (a)(i) to write down an estimate for .
Use the Lagrange form of the error term to find an upper bound for the absolute value of the error in calculating , using the first three non-zero terms of the Maclaurin series for .
With reference to the Lagrange form of the error term, explain whether your answer to part (b) is an overestimate or an underestimate for .
Markscheme
substitution of in (M1)
A1
[2 marks]
(M1)
Note: Award (M1) if this is seen in part (a)(i).
attempt to differentiate their answer in part (a) (M1)
M1
Note: Award M1 for equating their derivatives.
A1
[4 marks]
A1
Note: Accept an answer that rounds correct to or better.
[1 mark]
attempt to find the maximum of for (M1)
maximum of occurs at (A1)
(for all ) (A1)
use of (M1)
substitution of and and their value of and their value of into Lagrange error term (M1)
Note: Award (M1) for substitution of and and their value of and their value of into Lagrange error term.
upper bound A1
Note: Accept an answer that rounds correct to or better.
[6 marks]
(for all ) R1
Note: Accept or “the error term is negative”.
the answer in (b) is an overestimate A1
Note: The A1 is dependent on the R1.
[2 marks]
Examiners report
Question
This question asks you to explore the behaviour and key features of cubic polynomials of the form .
Consider the function for and where is a parameter, .
The graphs of for and are shown in the following diagrams.
On separate axes, sketch the graph of showing the value of the -intercept and the coordinates of any points with zero gradient, for
Hence, or otherwise, find the set of values of such that the graph of has
Given that the graph of has one local maximum point and one local minimum point, show that
Hence, for , find the set of values of such that the graph of has
.
.
Write down an expression for .
a point of inflexion with zero gradient.
one local maximum point and one local minimum point.
no points where the gradient is equal to zero.
the -coordinate of the local maximum point is .
the -coordinate of the local minimum point is .
exactly one -axis intercept.
exactly two -axis intercepts.
exactly three -axis intercepts.
Consider the function for and where .
Find all conditions on and such that the graph of has exactly one -axis intercept, explaining your reasoning.
Markscheme
: positive cubic with correct -intercept labelled A1
local maximum point correctly labelled A1
local minimum point correctly labelled A1
[3 marks]
: positive cubic with correct -intercept labelled A1
local maximum point correctly labelled A1
local minimum point correctly labelled A1
Note: Accept the following exact answers:
Local maximum point coordinates .
Local minimum point coordinates .
[3 marks]
A1
Note: Accept (an expression).
[1 mark]
A1
[1 mark]
considers the number of solutions to their (M1)
A1
[2 marks]
A1
Note: The (M1) in part (c)(ii) can be awarded for work shown in either (ii) or (iii).
[1 mark]
attempts to solve their for (M1)
(A1)
Note: Award (A1) if either or is subsequently considered.
Award the above (M1)(A1) if this work is seen in part (c).
correctly evaluates A1
the -coordinate of the local maximum point is AG
[3 marks]
correctly evaluates A1
the -coordinate of the local minimum point is AG
[1 mark]
the graph of will have one -axis intercept if
EITHER
(or equivalent reasoning) R1
OR
the minimum point is above the -axis R1
Note: Award R1 for a rigorous approach that does not (only) refer to sketched graphs.
THEN
A1
Note: Condone . The A1 is independent of the R1.
[2 marks]
the graph of will have two -axis intercepts if
EITHER
(or equivalent reasoning) (M1)
OR
evidence from the graph in part(a)(i) (M1)
THEN
A1
[2 marks]
the graph of will have three -axis intercepts if
EITHER
(or equivalent reasoning) (M1)
OR
reasoning from the results in both parts (e)(i) and (e)(ii) (M1)
THEN
A1
[2 marks]
case 1:
(independent of the value of ) A1
EITHER
does not have two solutions (has no solutions or solution) R1
OR
for R1
OR
the graph of has no local maximum or local minimum points, hence any vertical translation of this graph () will also have no local maximum or local minimum points R1
THEN
therefore there is only one -axis intercept AG
Note: Award at most A0R1 if only is considered.
case 2
is a local maximum point and is a local minimum point (A1)
Note: Award (A1) for a correct -coordinate seen for either the maximum or the minimum.
considers the positions of the local maximum point and/or the local minimum point (M1)
EITHER
considers both points above the -axis or both points below the -axis
OR
considers either the local minimum point only above the -axis OR the local maximum point only below the -axis
THEN
(both points above the -axis) A1
(both points above the -axis) A1
Note: Award at most (A1)(M1)A0A0 for case 2 if is not clearly stated.
[6 marks]
Examiners report
Question
This question asks you to explore the behaviour and some key features of the function , where and .
In parts (a) and (b), only consider the case where .
Consider .
Consider , where .
Now consider where and .
By using the result from part (f) and considering the sign of , show that the point on the graph of is
Sketch the graph of , stating the values of any axes intercepts and the coordinates of any local maximum or minimum points.
Use your graphic display calculator to explore the graph of for
• the odd values and ;
• the even values and .
Hence, copy and complete the following table.
Show that .
State the three solutions to the equation .
Show that the point on the graph of is always above the horizontal axis.
Hence, or otherwise, show that , for .
a local minimum point for even values of , where and .
a point of inflexion with zero gradient for odd values of , where and .
Consider the graph of , where , and .
State the conditions on and such that the equation has four solutions for .
Markscheme
inverted parabola extended below the -axis A1
-axis intercept values A1
Note: Accept a graph passing through the origin as an indication of .
local maximum at A1
Note: Coordinates must be stated to gain the final A1.
Do not accept decimal approximations.
[3 marks]
A1A1A1A1A1A1
Note: Award A1 for each correct value.
For a table not sufficiently or clearly labelled, assume that their values are in the same order as the table in the question paper and award marks accordingly.
[6 marks]
METHOD 1
attempts to use the product rule (M1)
A1A1
Note: Award A1 for a correct and A1 for a correct .
EITHER
attempts to factorise (involving at least one of or ) (M1)
A1
OR
attempts to express as the difference of two products with each product containing at least one of or (M1)
A1
THEN
AG
Note: Award the final (M1)A1 for obtaining any of the following forms:
METHOD 2
(M1)
A1
attempts to use the chain rule (M1)
A1A1
Note: Award A1 for and A1 for .
AG
[5 marks]
A2
Note: Award A1 for either two correct solutions or for obtaining
Award A0 otherwise.
[2 marks]
attempts to find an expression for (M1)
A1
EITHER
since (for and so ) R1
Note: Accept any logically equivalent conditions/statements on and .
Award R0 if any conditions/statements specified involving , or both are incorrect.
OR
(since ), raised to an even power () (or equivalent reasoning) is always positive (and so ) R1
Note: The condition is given in the question. Hence some candidates will assume and not state it. In these instances, award R1 for a convincing argument.
Accept any logically equivalent conditions/statements on on and .
Award R0 if any conditions/statements specified involving , or both are incorrect.
THEN
so is always above the horizontal axis AG
Note: Do not award (M1)A0R1.
[3 marks]
METHOD 1
A1
EITHER
as and R1
OR
and are all R1
Note: Do not award A0R1.
Accept equivalent reasoning on correct alternative expressions for and accept any logically equivalent conditions/statements on and .
Exceptions to the above are condone and condone .
An alternative form for is .
THEN
hence AG
METHOD 2
and A1
(since is continuous and there are no stationary points between and )
the gradient (of the curve) must be positive between and R1
Note: Do not award A0R1.
hence AG
[2 marks]
for even:
(and are both ) R1
A1
and (seen anywhere) A1
Note: Candidates can give arguments based on the sign of to obtain the R mark.
For example, award R1 for the following:
If is even, then is odd and hence .
Do not award R0A1.
The second A1 is independent of the other two marks.
The A marks can be awarded for correct descriptions expressed in words.
Candidates can state as a point of zero gradient from part (d) or show, state or explain (words or diagram) that . The last A mark can be awarded for a clearly labelled diagram showing changes in the sign of the gradient.
The last A1 can be awarded for use of a specific case (e.g. ).
hence is a local minimum point AG
[3 marks]
for odd:
, (and are both ) so R1
Note: Candidates can give arguments based on the sign of to obtain the R mark.
For example, award R1 for the following:
If is odd, then is even and hence .
and (seen anywhere) A1
Note: The A1 is independent of the R1.
Candidates can state as a point of zero gradient from part (d) or show, state or explain (words or diagram) that . The last A mark can be awarded for a clearly labelled diagram showing changes in the sign of the gradient.
The last A1 can be awarded for use of a specific case (e.g. ).
hence is a point of inflexion with zero gradient AG
[2 marks]
considers the parity of (M1)
Note: Award M1 for stating at least one specific even value of .
must be even (for four solutions) A1
Note: The above 2 marks are independent of the 3 marks below.
A1A1A1
Note: Award A1 for the correct lower endpoint, A1 for the correct upper endpoint and A1 for strict inequality signs.
The third A1 (strict inequality signs) can only be awarded if A1A1 has been awarded.
For example, award A1A1A0 for . Award A1A0A0 for .
Award A1A0A0 for .
[5 marks]
Examiners report
Question
This question asks you to examine various polygons for which the numerical value of the area is the same as the numerical value of the perimeter. For example, a by rectangle has an area of and a perimeter of .
For each polygon in this question, let the numerical value of its area be and let the numerical value of its perimeter be .
An -sided regular polygon can be divided into congruent isosceles triangles. Let be the length of each of the two equal sides of one such isosceles triangle and let be the length of the third side. The included angle between the two equal sides has magnitude .
Part of such an -sided regular polygon is shown in the following diagram.
Consider a -sided regular polygon such that .
The Maclaurin series for is
Consider a right-angled triangle with side lengths and , where , such that .
Find the side length, , where , of a square such that .
Write down, in terms of and , an expression for the area, , of one of these isosceles triangles.
Show that .
Use the results from parts (b) and (c) to show that .
Use the Maclaurin series for to find .
Interpret your answer to part (e)(i) geometrically.
Show that .
By using the result of part (f) or otherwise, determine the three side lengths of the only two right-angled triangles for which .
Determine the area and perimeter of these two right-angled triangles.
Markscheme
and (A1)
(M1)
A1
Note: Award A1M1A0 if both and are stated as final answers.
[3 marks]
A1
Note: Award A1 for a correct alternative form expressed in terms of and only.
For example, using Pythagoras’ theorem, or or .
[1 mark]
METHOD 1
uses (M1)
A1
AG
METHOD 2
uses Pythagoras’ theorem and (M1)
A1
AG
METHOD 3
uses the cosine rule (M1)
A1
AG
METHOD 4
uses the sine rule (M1)
A1
AG
[2 marks]
(M1)
Note: Award M1 for equating correct expressions for and .
A1
uses (seen anywhere in part (d) or in part (b)) (M1)
attempts to either factorise or divide their expression (M1)
(or equivalent) A1
EITHER
substitutes (or equivalent) into (M1)
A1
Note: Other approaches are possible. For example, award A1 for and M1 for substituting into .
OR
substitutes (or equivalent) into (M1)
A1
THEN
AG
[7 marks]
attempts to use the Maclaurin series for with (M1)
(or equivalent) A1
A1
Note: Award a maximum of M1A1A0 if is not stated anywhere.
[3 marks]
(as and )
the polygon becomes a circle of radius R1
Note: Award R1 for alternative responses such as:
the polygon becomes a circle of area OR
the polygon becomes a circle of perimeter OR
the polygon becomes a circle with .
Award R0 for polygon becomes a circle.
[1 mark]
and (A1)(A1)
equates their expressions for and M1
M1
Note: Award M1 for isolating or . This step may be seen later.
M1
Note: Award M1 for attempting to expand their RHS of either or .
EITHER
A1
OR
A1
THEN
A1
AG
Note: Award a maximum of A1A1M1M1M0A0A0 for attempting to verify.
For example, verifying that gains of the marks.
[7 marks]
using an appropriate method (M1)
eg substituting values for or using divisibility properties
and A1A1
Note: Award A1A0 for either one set of three correct side lengths or two sets of two correct side lengths.
[3 marks]
and A1
Note: Do not award A1FT.
[1 mark]
Examiners report
Question
In this question you will explore some of the properties of special functions and and their relationship with the trigonometric functions, sine and cosine.
Functions and are defined as and , where .
Consider and , such that .
Using , find expressions, in terms of and , for
The functions and are known as circular functions as the general point () defines points on the unit circle with equation .
The functions and are known as hyperbolic functions, as the general point ( ) defines points on a curve known as a hyperbola with equation . This hyperbola has two asymptotes.
Verify that satisfies the differential equation .
Show that .
.
.
Hence find, and simplify, an expression for .
Show that .
Sketch the graph of , stating the coordinates of any axis intercepts and the equation of each asymptote.
The hyperbola with equation can be rotated to coincide with the curve defined by .
Find the possible values of .
Markscheme
A1
A1
AG
[2 marks]
METHOD 1
substituting and M1
(M1)
A1
AG
METHOD 2
M1
M1A1
AG
Note: Accept combinations of METHODS 1 & 2 that meet at equivalent expressions.
[3 marks]
substituting into the expression for (M1)
obtaining (A1)
Note: The M1 can be awarded for the use of sine and cosine being odd and even respectively.
A1
[3 marks]
substituting and attempt to simplify (M1)
A1
[2 marks]
METHOD 1
substituting expressions found in part (c) (M1)
A1
METHOD 2
M1
A1
Note: Accept equivalent final answers that have been simplified removing all imaginary parts eg etc
[2 marks]
M1
A1
A1
Note: Award A1 for a value of 1 obtained from either LHS or RHS of given expression.
M1
(hence ) AG
Note: Award full marks for showing that .
[4 marks]
A1A1A1A1
Note: Award A1 for correct curves in the upper quadrants, A1 for correct curves in the lower quadrants, A1 for correct -intercepts of and (condone and ), A1 for and .
[4 marks]
attempt to rotate by in either direction (M1)
Note: Evidence of an attempt to relate to a sketch of would be sufficient for this (M1).
attempting to rotate a particular point, eg (M1)
rotates to (or similar) (A1)
hence A1A1
[5 marks]
Examiners report
Question
In this question you will be exploring the strategies required to solve a system of linear differential equations.
Consider the system of linear differential equations of the form:
and ,
where and is a parameter.
First consider the case where .
Now consider the case where .
Now consider the case where .
From previous cases, we might conjecture that a solution to this differential equation is , and is a constant.
By solving the differential equation , show that where is a constant.
Show that .
Solve the differential equation in part (a)(ii) to find as a function of .
By differentiating with respect to , show that .
By substituting , show that where is a constant.
Hence find as a function of .
Hence show that , where is a constant.
Show that .
Find the two values for that satisfy .
Let the two values found in part (c)(ii) be and .
Verify that is a solution to the differential equation in (c)(i),where is a constant.
Markscheme
METHOD 1
(M1)
OR A1A1
Note: Award A1 for and A1 for and .
AG
METHOD 2
rearranging to AND multiplying by integrating factor M1
A1A1
AG
[3 marks]
substituting into differential equation in M1
AG
[1 mark]
integrating factor (IF) is (M1)
(A1)
(A1)
A1
Note: The first constant must be , and the second can be any constant for the final A1 to be awarded. Accept a change of constant applied at the end.
[4 marks]
A1
EITHER
(M1)
A1
OR
(M1)
A1
THEN
AG
[3 marks]
A1
M1
OR A1
AG
[3 marks]
M1
A1
Note: The first constant must be , and the second can be any constant for the final A1 to be awarded. Accept a change of constant applied at the end.
[2 marks]
METHOD 1
substituting and their (iii) into M1(M1)
A1
AG
Note: Follow through from incorrect part (iii) cannot be awarded if it does not lead to the AG.
METHOD 2
M1
A1
M1
AG
[3 marks]
seen anywhere M1
METHOD 1
attempt to eliminate M1
A1
AG
METHOD 2
rewriting LHS in terms of and M1
A1
AG
[3 marks]
(A1)
(M1)
(since ) A1
and are and (either order) A1
[4 marks]
METHOD 1
(A1)(A1)
M1
A1
AG
METHOD 2
(A1)(A1)
M1
A1
AG
[4 marks]
Examiners report
Question
This question asks you to explore properties of a family of curves of the type for various values of and , where .
On the same set of axes, sketch the following curves for and , clearly indicating any points of intersection with the coordinate axes.
Now, consider curves of the form , for , where .
Next, consider the curve .
The curve has two points of inflexion. Due to the symmetry of the curve these points have the same -coordinate.
is defined to be a rational point on a curve if and are rational numbers.
The tangent to the curve at a rational point intersects the curve at another rational point .
Let be the curve , for . The rational point lies on .
Write down the coordinates of the two points of inflexion on the curve .
By considering each curve from part (a), identify two key features that would distinguish one curve from the other.
By varying the value of , suggest two key features common to these curves.
Show that , for .
Hence deduce that the curve has no local minimum or maximum points.
Find the value of this -coordinate, giving your answer in the form , where .
Find the equation of the tangent to at .
Hence, find the coordinates of the rational point where this tangent intersects , expressing each coordinate as a fraction.
The point also lies on . The line intersects at a further point. Determine the coordinates of this point.
Markscheme
approximately symmetric about the -axis graph of A1
including cusp/sharp point at A1
[2 marks]
Note: Final A1 can be awarded if intersections are in approximate correct place with respect to the axes shown. Award A1A1A1A0 if graphs ‘merge’ or ‘cross’ or are discontinuous at -axis but are otherwise correct. Award A1A0A0A0 if only one correct branch of both curves are seen.
Note: If they sketch graphs on separate axes, award a maximum of 2 marks for the ‘best’ response seen. This is likely to be A1A1A0A0.
approximately symmetric about the -axis graph of with approximately correct gradient at axes intercepts A1
some indication of position of intersections at , A1
[2 marks]
Note: Final A1 can be awarded if intersections are in approximate correct place with respect to the axes shown. Award A1A1A1A0 if graphs ‘merge’ or ‘cross’ or are discontinuous at -axis but are otherwise correct. Award A1A0A0A0 if only one correct branch of both curves are seen.
Note: If they sketch graphs on separate axes, award a maximum of 2 marks for the ‘best’ response seen. This is likely to be A1A1A0A0.
and A1
[1 mark]
Any two from:
has a cusp/sharp point, (the other does not)
graphs have different domains
has points of inflexion, (the other does not)
graphs have different -axis intercepts (one goes through the origin, and the other does not)
graphs have different -axis intercepts A1
Note: Follow through from their sketch in part (a)(i). In accordance with marking rules, mark their first two responses and ignore any subsequent.
[1 mark]
Any two from:
as ,
as is approximated by (or similar)
they have intercepts at
they have intercepts at
they all have the same range
(or -axis) is a line of symmetry
they all have the same line of symmetry
they have one -axis intercept
they have two -axis intercepts
they have two points of inflexion
at -axis intercepts, curve is vertical/infinite gradient
there is no cusp/sharp point at -axis intercepts A1A1
Note: The last example is the only valid answer for things “not” present. Do not credit an answer of “they are all symmetrical” without some reference to the line of symmetry.
Note: Do not allow same/ similar shape or equivalent.
Note: In accordance with marking rules, mark their first two responses and ignore any subsequent.
[2 marks]
METHOD 1
attempt to differentiate implicitly M1
A1
OR A1
AG
METHOD 2
attempt to use chain rule M1
A1A1
Note: Award A1 for , A1 for
AG
[3 marks]
EITHER
local minima/maxima occur when
has no (real) solutions (or equivalent) R1
OR
, so R1
THEN
so, no local minima/maxima exist AG
[1 mark]
EITHER
attempt to use quotient rule to find M1
A1A1
Note: Award A1 for correct and correct denominator, A1 for correct .
Note: Future A marks may be awarded if the denominator is missing or incorrect.
stating or using (may be seen anywhere) (M1)
OR
attempt to use product rule to find M1
A1A1
Note: Award A1 for correct first term, A1 for correct second term.
setting (M1)
OR
attempts implicit differentiation on M1
A1
recognizes that (M1)
(A1)
THEN
A1
attempt to use quadratic formula or equivalent (M1)
A1
Note: Accept any integer multiple of and (e.g. and ).
[7 marks]
attempt to find tangent line through (M1)
OR A1
[2 marks]
attempt to solve simultaneously with (M1)
Note: The M1 mark can be awarded for an unsupported correct answer in an incorrect format (e.g. ).
obtain A1
[2 marks]
attempt to find equation of (M1)
(A1)
solve simultaneously with (M1)
A1
A1
OR
attempt to find vector equation of (M1)
(A1)
attempt to solve (M1)
A1
A1
[5 marks]
Examiners report
This was a relatively straightforward start, though it was disappointing to see so many candidates sketch their graphs on two separate axes, despite the question stating they should be sketched on the same axes.
Of those candidates producing clear sketches, the vast majority were able to recognise the points of inflexion and write down their coordinates. A small number embarked on a mostly fruitless algebraic approach rather than use their graphs as intended. The distinguishing features between curves tended to focus on points of intersection with the axes, which was accepted. Only a small number offered ideas such as on both curves. A number of (incorrect) suggestions were seen, stating that both curves tended towards a linear asymptote.
A majority of candidates' suggestions related to the number of intersection points with the coordinate axes, while the idea of the x-axis acting as a line of symmetry was also often seen.
The required differentiation was straightforward for the majority of candidates.
The majority employed the quotient rule here, often doing so successfully to find a correct expression for . Despite realising that , the resulting algebra to find the required solution proved a step too far for most. A number of slips were seen in candidates' working, though better candidates were able to answer the question confidently.
Mistakes proved to be increasingly common by this stage of the paper. Various equations of lines were suggested, with the incorrect appearing more than once. Only the better candidates were able to tackle the final part of the question with any success; it was pleasing to see a number of clear algebraic (only) approaches, though this was not necessary to obtain full marks.
Significant work on this question part was rarely seen, and it may have been the case that many candidates chose to spend their remaining time on the second question, especially if they felt they were making little progress with part f. Having said that, correct final answers were seen from better candidates, though these were few and far between.
Question
This question asks you to explore cubic polynomials of the form for and corresponding cubic equations with one real root and two complex roots of the form for .
In parts (a), (b) and (c), let and .
Consider the equation for .
Consider the function for .
Consider the function for where and .
The equation for has roots and where and .
On the Cartesian plane, the points and represent the real and imaginary parts of the complex roots of the equation .
The following diagram shows a particular curve of the form and the tangent to the curve at the point . The curve and the tangent both intersect the -axis at the point . The points and are also shown.
Consider the curve for . The points and are as defined in part (d)(ii). The curve has a point of inflexion at point .
Consider the special case where and .
Given that and are roots of the equation, write down the third root.
Verify that the mean of the two complex roots is .
Show that the line is tangent to the curve at the point .
Sketch the curve and the tangent to the curve at point , clearly showing where the tangent crosses the -axis.
Show that .
Hence, or otherwise, prove that the tangent to the curve at the point intersects the -axis at the point .
Deduce from part (d)(i) that the complex roots of the equation can be expressed as .
Use this diagram to determine the roots of the corresponding equation of the form for .
State the coordinates of .
Show that the -coordinate of is .
You are not required to demonstrate a change in concavity.
Hence describe numerically the horizontal position of point relative to the horizontal positions of the points and .
Sketch the curve for and .
For and , state in terms of , the coordinates of points and .
Markscheme
A1
[1 mark]
mean A1
AG
[1 mark]
METHOD 1
attempts product rule differentiation (M1)
Note: Award (M1) for attempting to express as
A1
A1
Note: Where is correct, award A1 for solving and obtaining .
EITHER
A1
OR
A1
OR
states the gradient of is also and verifies that lies on the line A1
THEN
so is the tangent to the curve at AG
Note: Award a maximum of (M0)A0A1A1 to a candidate who does not attempt to find .
METHOD 2
sets to form (M1)
EITHER
A1
attempts to solve a correct cubic equation (M1)
OR
recognises that and forms A1
attempts to solve a correct quadratic equation (M1)
THEN
is a double root R1
so is the tangent to the curve at AG
Note: Candidates using this method are not required to verify that .
[4 marks]
a positive cubic with an -intercept , and a local maximum and local minimum in the first quadrant both positioned to the left of A1
Note: As the local minimum and point A are very close to each other, condone graphs that seem to show these points coinciding.
For the point of tangency, accept labels such as or the point labelled from both axes. Coordinates are not required.
a correct sketch of the tangent passing through and crossing the -axis at the same point as the curve A1
Note: Award A1A0 if both graphs cross the -axis at distinctly different points.
[2 marks]
EITHER
(M1)A1
OR
attempts to find M1
A1
THEN
AG
[2 marks]
METHOD 1
(A1)
(A1)
attempts to substitute their and into M1
EITHER
A1
sets so M1
OR R1
OR
sets so M1
OR R1
A1
THEN
so the tangent intersects the -axis at the point AG
METHOD 2
(A1)
(A1)
attempts to substitute their and into and attempts to find M1
EITHER
A1
sets so M1
OR R1
OR
sets so M1
OR R1
A1
METHOD 3
(A1)
the line through parallel to the tangent at has equation
A1
sets to form M1
A1
A1
since there is a double root , this parallel line through is the required tangent at R1
[6 marks]
EITHER
(since ) R1
Note: Accept .
OR
and R1
THEN
hence the complex roots can be expressed as AG
[1 mark]
(seen anywhere) A1
EITHER
attempts to find the gradient of the tangent in terms of and equates to (M1)
OR
substitutes and to form (M1)
OR
substitutes and into (M1)
THEN
roots are (seen anywhere) and A1A1
Note: Award A1 for and A1 for . Do not accept coordinates.
[4 marks]
A1
Note: Accept “ and ”.
Do not award A1FT for .
[1 mark]
attempts to find M1
sets and correctly solves for A1
for example, obtaining leading to
so AG
Note: Do not award A1 if the answer does not lead to the AG.
[2 marks]
point is of the horizontal distance (way) from point to point A1
Note: Accept equivalent numerical statements or a clearly labelled diagram displaying the numerical relationship.
Award A0 for non-numerical statements such as “ is between and , closer to ”.
[1 mark]
(A1)
a positive cubic with no stationary points and a non-stationary point of inflexion at A1
Note: Graphs may appear approximately linear. Award this A1 if a change of concavity either side of is apparent.
Coordinates are not required and the -intercept need not be indicated.
[2 marks]
A1
[1 mark]
Examiners report
Part (a) (i) was generally well done with a significant majority of candidates using the conjugate root theorem to state as the third root. A number of candidates, however, wasted considerable time attempting an algebraic method to determine the third root. Part (a) (ii) was reasonably well done. A few candidates however attempted to calculate the product of and .
Part (b) was reasonably well done by a significant number of candidates. Most were able to find a correct expression for and a good number of those candidates were able to determine that . Candidates that did not determine the equation of the tangent had to state that the gradient of is also 1 and verify that the point (4,3) lies on the line. A few candidates only met one of those requirements. Weaker candidates tended to only verify that the point (4,3) lies on the curve and the tangent line without attempting to find .
Part (c) was not answered as well as anticipated. A number of sketches were inaccurate and carelessly drawn with many showing both graphs crossing the x-axis at distinctly different points.
Part (d) (i) was reasonably well done by a good number of candidates. Most successful responses involved use of the product rule. A few candidates obtained full marks by firstly expanding , then differentiating to find and finally simplifying to obtain the desired result. A number of candidates made elementary mistakes when differentiating. In general, the better candidates offered reasonable attempts at showing the general result in part (d) (ii). A good number gained partial credit by determining that and/or . Only the very best candidates obtained full marks by concluding that as or , then when .
In general, only the best candidates were able to use the result to deduce that the complex roots of the equation can be expressed as . Although given the complex roots , a significant number of candidates attempted, with mixed success, to use the quadratic formula to solve the equation .
In part (f) (i), only a small number of candidates were able to determine all the roots of the equation. Disappointingly, a large number did not state as a root. Some candidates determined that but were unable to use the diagram to determine that . Of the candidates who determined all the roots in part (f) (i), very few gave the correct coordinates for C2 . The most frequent error was to give the y-coordinate as .
Of the candidates who attempted part (g) (i), most were able to find an expression for and a reasonable number of these were then able to convincingly show that . It was very rare to see a correct response to part (g) (ii). A few candidates stated that P is between R and A with some stating that P was closer to A. A small number restated in words.
Of the candidates who attempted part (h) (i), most were able to determine that . However, most graphs were poorly drawn with many showing a change in concavity at rather than at . In part (h) (ii), only a very small number of candidates determined that A and P coincide at (r,0).
Question
This question will investigate power series, as an extension to the Binomial Theorem for negative and fractional indices.
A power series in is defined as a function of the form where the .
It can be considered as an infinite polynomial.
This is an example of a power series, but is only a finite power series, since only a finite number of the are non-zero.
We will now attempt to generalise further.
Suppose can be written as the power series .
Expand using the Binomial Theorem.
Consider the power series
By considering the ratio of consecutive terms, explain why this series is equal to and state the values of for which this equality is true.
Differentiate the equation obtained part (b) and hence, find the first four terms in a power series for .
Repeat this process to find the first four terms in a power series for .
Hence, by recognising the pattern, deduce the first four terms in a power series for , .
By substituting , find the value of .
By differentiating both sides of the expression and then substituting , find the value of .
Repeat this procedure to find and .
Hence, write down the first four terms in what is called the Extended Binomial Theorem for .
Write down the power series for .
Hence, using integration, find the power series for , giving the first four non-zero terms.
Markscheme
M1A1
[2 marks]
It is an infinite GP with R1A1
M1A1AG
[4 marks]
A1
A1
[2 marks]
A1
A1
[2 marks]
A1A1A1
[3 marks]
A1
[1 mark]
A1
A1
[2 marks]
A1
A1
A1
A1
[4 marks]
A1
[1 mark]
M1A1
[2 marks]
M1A1
Putting R1
So A1
[4 marks]
Examiners report
Question
This question will investigate methods for finding definite integrals of powers of trigonometrical functions.
Let .
Let
Let .
Find the exact values of , and .
Use integration by parts to show that .
Explain where the condition was used in your proof.
Hence, find the exact values of and .
Use the substitution to show that .
Hence, find the exact values of and
Find the exact values of and .
Use the fact that to show that .
Explain where the condition was used in your proof.
Hence, find the exact values of and .
Markscheme
M1A1
M1A1
M1A1
[6 marks]
M1A1A1
M1A1
AG
[6 marks]
need so that in R1
[1 mark]
A1A1
[2 marks]
A1
M1A1A1AG
[4 marks]
A1A1
[2 marks]
A1
M1A1
[3 marks]
M1
A1A1AG
[3 marks]
need so that the powers of tan in are not negative R1
[1 mark]
A1
A1
[2 marks]
Examiners report
Question
This question investigates some applications of differential equations to modeling population growth.
One model for population growth is to assume that the rate of change of the population is proportional to the population, i.e. , where , is the time (in years) and is the population
The initial population is 1000.
Given that , use your answer from part (a) to find
Consider now the situation when is not a constant, but a function of time.
Given that , find
Another model for population growth assumes
- there is a maximum value for the population, .
- that is not a constant, but is proportional to .
Show that the general solution of this differential equation is , where .
the population after 10 years
the number of years it will take for the population to triple.
the solution of the differential equation, giving your answer in the form .
the number of years it will take for the population to triple.
Show that , where .
Solve the differential equation , giving your answer in the form .
Given that the initial population is 1000, and , find the number of years it will take for the population to triple.
Markscheme
M1A1
A1A1
A1
, where AG
[5 marks]
when
A1
A1
[2 marks]
M1
years A1
[2 marks]
A1
[1 mark]
M1
A1A1
A1
when
M1
[5 marks]
M1
A1
Use of quadratic formula or GDC graph or GDC polysmlt M1
years A1
[4 marks]
, where is the constant of proportionality A1
So A1
AG
[2 marks]
M1
M1
A1
A1
A1A1
, where M1
, where A1
M1
A1
[10 marks]
M1
A1
M1
years A1
[4 marks]
Examiners report
Question
This question asks you to investigate regular -sided polygons inscribed and circumscribed in a circle, and the perimeter of these as tends to infinity, to make an approximation for .
Let represent the perimeter of any -sided regular polygon inscribed in a circle of radius 1 unit.
Consider an equilateral triangle ABC of side length, units, circumscribed about a circle of radius 1 unit and centre O as shown in the following diagram.
Let represent the perimeter of any -sided regular polygon circumscribed about a circle of radius 1 unit.
Consider an equilateral triangle ABC of side length, units, inscribed in a circle of radius 1 unit and centre O as shown in the following diagram.
The equilateral triangle ABC can be divided into three smaller isosceles triangles, each subtending an angle of at O, as shown in the following diagram.
Using right-angled trigonometry or otherwise, show that the perimeter of the equilateral triangle ABC is equal to units.
Consider a square of side length, units, inscribed in a circle of radius 1 unit. By dividing the inscribed square into four isosceles triangles, find the exact perimeter of the inscribed square.
Find the perimeter of a regular hexagon, of side length, units, inscribed in a circle of radius 1 unit.
Show that .
Use an appropriate Maclaurin series expansion to find and interpret this result geometrically.
Show that .
By writing in the form , find .
Use the results from part (d) and part (f) to determine an inequality for the value of in terms of .
The inequality found in part (h) can be used to determine lower and upper bound approximations for the value of .
Determine the least value for such that the lower bound and upper bound approximations are both within 0.005 of .
Markscheme
METHOD 1
consider right-angled triangle OCX where CX
M1A1
A1
AG
METHOD 2
eg use of the cosine rule M1A1
A1
AG
Note: Accept use of sine rule.
[3 marks]
where = side of square M1
A1
A1
[3 marks]
6 equilateral triangles ⇒ = 1 A1
A1
[2 marks]
in right-angled triangle M1
A1
M1
AG
[3 marks]
consider
use of M1
(A1)
A1
A1
as polygon becomes a circle of radius 1 and R1
[5 marks]
consider an -sided polygon of side length
2 right-angled triangles with angle at centre M1A1
opposite side M1A1
Perimeter AG
[4 marks]
consider
R1
attempt to use L’Hopital’s rule M1
A1A1
A1
[5 marks]
M1
A1
[2 marks]
attempt to find the lower bound and upper bound approximations within 0.005 of (M1)
= 46 A2
[3 marks]
Examiners report
Question
This question asks you to investigate some properties of the sequence of functions of the form , −1 ≤ ≤ 1 and .
Important: When sketching graphs in this question, you are not required to find the coordinates of any axes intercepts or the coordinates of any stationary points unless requested.
For odd values of > 2, use your graphic display calculator to systematically vary the value of . Hence suggest an expression for odd values of describing, in terms of , the number of
For even values of > 2, use your graphic display calculator to systematically vary the value of . Hence suggest an expression for even values of describing, in terms of , the number of
The sequence of functions, , defined above can be expressed as a sequence of polynomials of degree .
Consider .
On the same set of axes, sketch the graphs of and for −1 ≤ ≤ 1.
local maximum points;
local minimum points;
On a new set of axes, sketch the graphs of and for −1 ≤ ≤ 1.
local maximum points;
local minimum points.
Solve the equation and hence show that the stationary points on the graph of occur at where and 0 < < .
Use an appropriate trigonometric identity to show that .
Use an appropriate trigonometric identity to show that .
Hence show that , .
Hence express as a cubic polynomial.
Markscheme
correct graph of A1
correct graph of A1
[2 marks]
graphical or tabular evidence that has been systematically varied M1
eg = 3, 1 local maximum point and 1 local minimum point
= 5, 2 local maximum points and 2 local minimum points
= 7, 3 local maximum points and 3 local minimum points (A1)
local maximum points A1
[3 marks]
local minimum points A1
Note: Allow follow through from an incorrect local maximum formula expression.
[1 mark]
correct graph of A1
correct graph of A1
[2 marks]
graphical or tabular evidence that has been systematically varied M1
eg = 2, 0 local maximum point and 1 local minimum point
= 4, 1 local maximum points and 2 local minimum points
= 6, 2 local maximum points and 3 local minimum points (A1)
local maximum points A1
[3 marks]
local minimum points A1
[1 mark]
M1A1
Note: Award M1 for attempting to use the chain rule.
M1
A1
leading to
( and 0 < < ) AG
[4 marks]
M1
stating that A1
so AG
[2 marks]
A1
use of cos(A + B) = cos A cos B − sin A sin B leading to M1
AG
[2 marks]
A1
M1
A1
AG
[3 marks]
(M1)
A1
[2 marks]
Examiners report
Question
Let . Find , given that .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
evidence of integration (M1)
eg
correct integration (accept missing ) (A2)
eg
substituting initial condition into their integrated expression (must have ) M1
eg
Note: Award M0 if they substitute into the original or differentiated function.
recognizing (A1)
eg
(A1)
A1 N5
[7 marks]
Examiners report
Question
Let .
Let , where .
Let and .
(i) Find the first four derivatives of .
(ii) Find .
(i) Find the first three derivatives of .
(ii) Given that , find .
(i) Find .
(ii) Hence, show that .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(i) A2 N2
(ii) valid approach (M1)
egrecognizing that 19 is one less than a multiple of 4,
A1 N2
[4 marks]
(i)
A1A1 N2
(ii) METHOD 1
correct working that leads to the correct answer, involving the correct expression for the 19th derivative A2
eg
(accept ) A1 N1
METHOD 2
correct working involving recognizing patterns in coefficients of first three derivatives (may be seen in part (b)(i)) leading to a general rule for 19th coefficient A2
eg
(accept ) A1 N1
[5 marks]
(i) valid approach using product rule (M1)
eg
correct 20th derivatives (must be seen in product rule) (A1)(A1)
eg
A1 N3
(ii) substituting (seen anywhere) (A1)
eg
evidence of one correct value for or (seen anywhere) (A1)
eg
evidence of correct values substituted into A1
eg
Note: If candidates write only the first line followed by the answer, award A1A0A0.
AG N0
[7 marks]
Examiners report
Question
The equation of a curve is .
The gradient of the tangent to the curve at a point P is .
Find .
Find the coordinates of P.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)(A1) (C2)
Note: Award (A1) for , award (A1) for .
Award at most (A1)(A0) if there are any extra terms.
[2 marks]
(M1)
Note: Award (M1) for equating their answer to part (a) to .
(A1)(ft)
Note: Follow through from part (a). Award (M0)(A0) for seen without working.
(M1)
Note: Award (M1) substituting their into the original function.
(A1)(ft) (C4)
Note: Accept .
[4 marks]
Examiners report
Question
Let . Given that , find .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach (M1)
eg
correct integration by substitution/inspection A2
eg
correct substitution into their integrated function (must include ) M1
eg
Note: Award M0 if candidates substitute into or .
(A1)
A1 N4
[6 marks]
Examiners report
Question
Find .
Find , given that and .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach to set up integration by substitution/inspection (M1)
eg
correct expression (A1)
eg
A2 N4
Notes: Award A1 if missing “”.
[4 marks]
substituting into their answer from (a) (M1)
eg
correct working (A1)
eg
A1 N2
[3 marks]
Examiners report
Question
A quadratic function can be written in the form . The graph of has axis of symmetry and -intercept at
Find the value of .
Find the value of .
The line is a tangent to the curve of . Find the values of .
Markscheme
METHOD 1 (using x-intercept)
determining that 3 is an -intercept (M1)
eg, 
valid approach (M1)
eg
A1 N2
METHOD 2 (expanding f (x))
correct expansion (accept absence of ) (A1)
eg
valid approach involving equation of axis of symmetry (M1)
eg
A1 N2
METHOD 3 (using derivative)
correct derivative (accept absence of ) (A1)
eg
valid approach (M1)
eg
A1 N2
[3 marks]
attempt to substitute (M1)
eg
correct working (A1)
eg
A1 N2
[3 marks]
METHOD 1 (using discriminant)
recognizing tangent intersects curve once (M1)
recognizing one solution when discriminant = 0 M1
attempt to set up equation (M1)
eg
rearranging their equation to equal zero (M1)
eg
correct discriminant (if seen explicitly, not just in quadratic formula) A1
eg
correct working (A1)
eg
A1A1 N0
METHOD 2 (using derivatives)
attempt to set up equation (M1)
eg
recognizing derivative/slope are equal (M1)
eg
correct derivative of (A1)
eg
attempt to set up equation in terms of either or M1
eg
rearranging their equation to equal zero (M1)
eg
correct working (A1)
eg
A1A1 N0
[8 marks]
Examiners report
Question
The following table shows the probability distribution of a discrete random variable , in terms of an angle .

Show that .
Given that , find .
Let , for . The graph of between and is rotated 360° about the -axis. Find the volume of the solid formed.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
evidence of summing to 1 (M1)
eg
correct equation A1
eg
correct equation in A1
eg
evidence of valid approach to solve quadratic (M1)
egfactorizing equation set equal to
correct working, clearly leading to required answer A1
eg
correct reason for rejecting R1
eg is a probability (value must lie between 0 and 1),
Note: Award R0 for without a reason.
AG N0
valid approach (M1)
egsketch of right triangle with sides 3 and 4,
correct working
(A1)
egmissing side
A1 N2
[3 marks]
attempt to substitute either limits or the function into formula involving (M1)
eg
correct substitution of both limits and function (A1)
eg
correct integration (A1)
eg
substituting their limits into their integrated function and subtracting (M1)
eg
Note: Award M0 if they substitute into original or differentiated function.
(A1)
eg
A1 N3
[6 marks]
Examiners report
Question
The diagram shows part of the graph of a function . The graph passes through point .

The tangent to the graph of at A has equation . Let be the normal to the graph of at A.
Write down the value of .
Find the equation of . Give your answer in the form where , , .
Draw the line on the diagram above.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
3 (A1) (C1)
Notes: Accept
[1 mark]
OR (A1)(A1)
Note: Award (A1) for correct gradient, (A1) for correct substitution of in the equation of line.
or any integer multiple (A1)(ft) (C3)
Note: Award (A1)(ft) for their equation correctly rearranged in the indicated form.
The candidate’s answer must be an equation for this mark.
[3 marks]
(M1)(A1)(ft) (C2)
Note: Award M1) for a straight line, with positive gradient, passing through , (A1)(ft) for line (or extension of their line) passing approximately through 2.5 or their intercept with the -axis.
[2 marks]
Examiners report
Question
Let , for . The following diagram shows part of the graph of and the rectangle OABC, where A is on the negative -axis, B is on the graph of , and C is on the -axis.

Find the -coordinate of A that gives the maximum area of OABC.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to find the area of OABC (M1)
eg
correct expression for area in one variable (A1)
eg
valid approach to find maximum area (seen anywhere) (M1)
eg
correct derivative A1
eg
correct working (A1)
eg
A2 N3
[7 marks]
Examiners report
Question
Consider , for , where .
The equation has exactly one solution. Find the value of .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1 – using discriminant
correct equation without logs (A1)
eg
valid approach (M1)
eg
recognizing discriminant must be zero (seen anywhere) M1
eg
correct discriminant (A1)
eg
correct working (A1)
eg
A2 N2
METHOD 2 – completing the square
correct equation without logs (A1)
eg
valid approach to complete the square (M1)
eg
correct working (A1)
eg
recognizing conditions for one solution M1
eg
correct working (A1)
eg
A2 N2
[7 marks]
Examiners report
Question
Let , for . The following diagram shows part of the graph of .

The graph of crosses the -axis at the origin and at the point .
The line intersects the graph of at another point Q, as shown in the following diagram.

Find the area of the region enclosed by the graph of and the line .
Markscheme
valid approach (M1)
eg, splitting area into triangles and integrals
correct integration (A1)(A1)
eg
substituting their limits into their integrated function and subtracting (in any order) (M1)
eg
Note: Award M0 for substituting into original or differentiated function.
area A2 N3
[6 marks]
Examiners report
Question
Rosewood College has 120 students. The students can join the sports club () and the music club ().
For a student chosen at random from these 120, the probability that they joined both clubs is and the probability that they joined the music club is.
There are 20 students that did not join either club.
Complete the Venn diagram for these students.

One of the students who joined the sports club is chosen at random. Find the probability that this student joined both clubs.
Determine whether the events and are independent.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)(A1) (C2)
Note: Award (A1) for 30 in correct area, (A1) for 60 and 10 in the correct areas.
[2 marks]
(A1)(ft)(A1)(ft) (C2)
Note: Award (A1)(ft) for correct numerator of 30, (A1)(ft) for correct denominator of 90. Follow through from their Venn diagram.
[2 marks]
(R1)
Note: Award (R1) for multiplying their by .
therefore the events are independent (A1)(ft) (C2)
Note: Award (R1)(A1)(ft) for an answer which is consistent with their Venn diagram.
Do not award (R0)(A1)(ft).
Do not award final (A1) if is not calculated. Follow through from part (a).
[2 marks]
Examiners report
Question
A quadratic function is given by . The points and lie on the graph of .
The -coordinate of the minimum of the graph is 3.
Find the equation of the axis of symmetry of the graph of .
Write down the value of .
Find the value of and of .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)(A1) (C2)
Note: Award (A1) for (a constant) and (A1) for .
[2 marks]
(A1) (C1)
[1 mark]
or equivalent
or equivalent
or equivalent (M1)
Note: Award (M1) for two of the above equations.
(A1)(ft)
(A1)(ft) (C3)
Note: Award at most (M1)(A1)(ft)(A0) if the answers are reversed.
Follow through from parts (a) and (b).
[3 marks]
Examiners report
Question
Maria owns a cheese factory. The amount of cheese, in kilograms, Maria sells in one week, , is given by
,
where is the price of a kilogram of cheese in euros (EUR).
Maria earns for each kilogram of cheese sold.
To calculate her weekly profit , in EUR, Maria multiplies the amount of cheese she sells by the amount she earns per kilogram.
Write down how many kilograms of cheese Maria sells in one week if the price of a kilogram of cheese is 8 EUR.
Find how much Maria earns in one week, from selling cheese, if the price of a kilogram of cheese is 8 EUR.
Write down an expression for in terms of .
Find the price, , that will give Maria the highest weekly profit.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
522 (kg) (A1) (C1)
[1 mark]
or equivalent (M1)
Note: Award (M1) for multiplying their answer to part (a) by .
626 (EUR) (626.40) (A1)(ft) (C2)
Note: Follow through from part (a).
[2 marks]
(A1)
OR
(A1) (C1)
[1 mark]
sketch of with some indication of the maximum (M1)
OR
(M1)
Note: Award (M1) for equating the correct derivative of their part (c) to zero.
OR
(M1)
Note: Award (M1) for correct substitution into the formula for axis of symmetry.
(A1)(ft) (C2)
Note: Follow through from their part (c), if the value of is such that .
[2 marks]
Examiners report
Question
Let , for .
Find .
Part of the graph of f is shown in the following diagram.
The shaded region R is enclosed by the graph of f, the x-axis, and the lines x = 1 and x = 9 . Find the volume of the solid formed when R is revolved 360° about the x-axis.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
correct working (A1)
eg
A2 N3
Note: Award A1 for .
[3 marks]
attempt to substitute either limits or the function into formula involving f 2 (accept absence of / dx) (M1)
eg
substituting limits into their integral and subtracting (in any order) (M1)
eg
correct working involving calculating a log value or using log law (A1)
eg
A1 N3
Note: Full FT may be awarded as normal, from their incorrect answer in part (a), however, do not award the final two A marks unless they involve logarithms.
[4 marks]
Examiners report
Question
Consider f(x), g(x) and h(x), for x∈ where h(x) = (x).
Given that g(3) = 7 , g′ (3) = 4 and f ′ (7) = −5 , find the gradient of the normal to the curve of h at x = 3.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
recognizing the need to find h′ (M1)
recognizing the need to find h′ (3) (seen anywhere) (M1)
evidence of choosing chain rule (M1)
eg
correct working (A1)
eg
(A1)
evidence of taking their negative reciprocal for normal (M1)
eg
gradient of normal is A1 N4
[7 marks]
Examiners report
Question
A closed cylindrical can with radius r centimetres and height h centimetres has a volume of 20 cm3.
The material for the base and top of the can costs 10 cents per cm2 and the material for the curved side costs 8 cents per cm2. The total cost of the material, in cents, is C.
Express h in terms of r.
Show that .
Given that there is a minimum value for C, find this minimum value in terms of .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
correct equation for volume (A1)
eg
A1 N2
[2 marks]
attempt to find formula for cost of parts (M1)
eg 10 × two circles, 8 × curved side
correct expression for cost of two circles in terms of r (seen anywhere) A1
eg
correct expression for cost of curved side (seen anywhere) (A1)
eg
correct expression for cost of curved side in terms of r A1
eg
AG N0
[4 marks]
recognize at minimum (R1)
eg
correct differentiation (may be seen in equation)
A1A1
correct equation A1
eg
correct working (A1)
eg
r = 2 (m) A1
attempt to substitute their value of r into C
eg (M1)
correct working
eg (A1)
(cents) A1 N3
Note: Do not accept 753.6, 753.98 or 754, even if 240 is seen.
[9 marks]
Examiners report
Question
The point A has coordinates (4 , −8) and the point B has coordinates (−2 , 4).
The point D has coordinates (−3 , 1).
Write down the coordinates of C, the midpoint of line segment AB.
Find the gradient of the line DC.
Find the equation of the line DC. Write your answer in the form ax + by + d = 0 where a , b and d are integers.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(1, −2) (A1)(A1) (C2)
Note: Award (A1) for 1 and (A1) for −2, seen as a coordinate pair.
Accept x = 1, y = −2. Award (A1)(A0) if x and y coordinates are reversed.
[2 marks]
(M1)
Note: Award (M1) for correct substitution, of their part (a), into gradient formula.
(A1)(ft) (C2)
Note: Follow through from part (a).
[2 marks]
OR OR (M1)
Note: Award (M1) for correct substitution of their part (b) and a given point.
OR
OR (M1)
Note: Award (M1) for correct substitution of their part (b) and a given point.
(accept any integer multiple, including negative multiples) (A1)(ft) (C2)
Note: Follow through from parts (a) and (b). Where the gradient in part (b) is found to be , award at most (M1)(A0) for either or .
[2 marks]
Examiners report
Question
In an international competition, participants can answer questions in only one of the three following languages: Portuguese, Mandarin or Hindi. 80 participants took part in the competition. The number of participants answering in Portuguese, Mandarin or Hindi is shown in the table.
A boy is chosen at random.
State the number of boys who answered questions in Portuguese.
Find the probability that the boy answered questions in Hindi.
Two girls are selected at random.
Calculate the probability that one girl answered questions in Mandarin and the other answered questions in Hindi.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
20 (A1) (C1)
[1 mark]
(A1)(A1) (C2)
Note: Award (A1) for correct numerator, (A1) for correct denominator.
[2 marks]
(A1)(M1)
Note: Award (A1) for first or second correct product seen, (M1) for adding their two products or for multiplying their product by two.
(A1) (C3)
[3 marks]
Examiners report
Question
A factory produces shirts. The cost, C, in Fijian dollars (FJD), of producing x shirts can be modelled by
C(x) = (x − 75)2 + 100.
The cost of production should not exceed 500 FJD. To do this the factory needs to produce at least 55 shirts and at most s shirts.
Find the cost of producing 70 shirts.
Find the value of s.
Find the number of shirts produced when the cost of production is lowest.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(70 − 75)2 + 100 (M1)
Note: Award (M1) for substituting in x = 70.
125 (A1) (C2)
[2 marks]
(s − 75)2 + 100 = 500 (M1)
Note: Award (M1) for equating C(x) to 500. Accept an inequality instead of =.
OR
(M1)
Note: Award (M1) for sketching correct graph(s).
(s =) 95 (A1) (C2)
[2 marks]
(M1)
Note: Award (M1) for an attempt at finding the minimum point using graph.
OR
(M1)
Note: Award (M1) for attempting to find the mid-point between their part (b) and 55.
OR
(C'(x) =) 2x − 150 = 0 (M1)
Note: Award (M1) for an attempt at differentiation that is correctly equated to zero.
75 (A1) (C2)
[2 marks]
Examiners report
Question
Let . The following diagram shows part of the graph of .
The region R is enclosed by the graph of , the -axis, and the -axis. Find the area of R.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1 (limits in terms of )
valid approach to find -intercept (M1)
eg , ,
-intercept is 3 (A1)
valid approach using substitution or inspection (M1)
eg , , , , ,
, , ,
(A2)
substituting both of their limits into their integrated function and subtracting (M1)
eg ,
Note: Award M0 if they substitute into original or differentiated function. Do not accept only “– 0” as evidence of substituting lower limit.
correct working (A1)
eg ,
area = 2 A1 N2
METHOD 2 (limits in terms of )
valid approach to find -intercept (M1)
eg , ,
-intercept is 3 (A1)
valid approach using substitution or inspection (M1)
eg , , , ,
, ,
correct integration (A2)
eg ,
both correct limits for (A1)
eg = 16 and = 25, , , = 4 and = 5, ,
substituting both of their limits for (do not accept 0 and 3) into their integrated function and subtracting (M1)
eg ,
Note: Award M0 if they substitute into original or differentiated function, or if they have not attempted to find limits for .
area = 2 A1 N2
[8 marks]
Examiners report
Question
Let . Part of the graph of is shown in the following diagram.
The graph of crosses the -axis at the point P. The line L is tangent to the graph of at P.
Find .
Hence, find the equation of L in terms of .
The graph of has a local minimum at the point Q. The line L passes through Q.
Find the value of .
Markscheme
A2 N2
[2 marks]
valid approach (M1)
eg
correct working (A1)
eg , slope = ,
attempt to substitute gradient and coordinates into linear equation (M1)
eg , , , L
correct equation A1 N3
eg , ,
[4 marks]
valid approach to find intersection (M1)
eg
correct equation (A1)
eg
correct working (A1)
eg ,
at Q (A1)
valid approach to find minimum (M1)
eg
correct equation (A1)
eg
substitution of their value of at Q into their equation (M1)
eg ,
= −4 A1 N0
[8 marks]
Examiners report
Question
Consider the curve y = 5x3 − 3x.
The curve has a tangent at the point P(−1, −2).
Find the equation of this tangent. Give your answer in the form y = mx + c.
Markscheme
(y − (−2)) = 12 (x − (−1)) (M1)
OR
−2 = 12(−1) + c (M1)
Note: Award (M1) for point and their gradient substituted into the equation of a line.
y = 12x + 10 (A1)(ft) (C2)
Note: Follow through from part (b).
[2 marks]
Examiners report
Question
The derivative of a function is given by . The graph of passes through .
Find .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
recognizing to integrate (M1)
eg , ,
correct integral (do not penalize for missing +) (A2)
eg
substituting (in any order) into their integrated expression (must have +) M1
eg
Note: Award M0 if they substitute into original or differentiated function.
(or any equivalent form, eg ) A1 N4
[5 marks]
Examiners report
Question
A particle P starts from point O and moves along a straight line. The graph of its velocity, ms−1 after seconds, for 0 ≤ ≤ 6 , is shown in the following diagram.
The graph of has -intercepts when = 0, 2 and 4.
The function represents the displacement of P from O after seconds.
It is known that P travels a distance of 15 metres in the first 2 seconds. It is also known that and .
Find the value of .
Find the total distance travelled in the first 5 seconds.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
recognizing relationship between and (M1)
eg ,
A1 N2
[2 marks]
correctly interpreting distance travelled in first 2 seconds (seen anywhere, including part (a) or the area of 15 indicated on diagram) (A1)
eg ,
valid approach to find total distance travelled (M1)
eg sum of 3 areas, , shaded areas in diagram between 0 and 5
Note: Award M0 if only is seen.
correct working towards finding distance travelled between 2 and 5 (seen anywhere including within total area expression or on diagram) (A1)
eg , , , ,
equal areas
correct working using (A1)
eg , , , ,
total distance travelled = 33 (m) A1 N2
[5 marks]
Examiners report
Question
Let be an obtuse angle such that .
Let .
Find the value of .
Line passes through the origin and has a gradient of . Find the equation of .
The following diagram shows the graph of for 0 ≤ ≤ 3. Line is a tangent to the graph of at point P.
Given that is parallel to , find the -coordinate of P.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
evidence of valid approach (M1)
eg sketch of triangle with sides 3 and 5,
correct working (A1)
eg missing side is 4 (may be seen in sketch), ,
A2 N4
[4 marks]
correct substitution of either gradient or origin into equation of line (A1)
(do not accept )
eg , ,
A2 N4
Note: Award A1A0 for .
[2 marks]
valid approach to equate their gradients (M1)
eg , , ,
correct equation without (A1)
eg , ,
correct working (A1)
eg ,
(do not accept ) A1 N1
Note: Do not award the final A1 if additional answers are given.
[4 marks]
Examiners report
Question
Let .
Consider the functions and , for ≥ 0.
The graphs of and are shown in the following diagram.
The shaded region is enclosed by the graphs of , , the -axis and .
Hence find .
Write down an expression for the area of .
Hence find the exact area of .
Markscheme
integrating by inspection from (a) or by substitution (M1)
eg , , , ,
correct integrated expression in terms of A2 N3
eg ,
[3 marks]
integrating and subtracting functions (in any order) (M1)
eg ,
correct integral (including limits, accept absence of ) A1 N2
eg , ,
[2 marks]
recognizing is a common factor (seen anywhere, may be seen in part (c)) (M1)
eg , ,
correct integration (A1)(A1)
eg
Note: Award A1 for and award A1 for .
substituting limits into their integrated function and subtracting (in any order) (M1)
eg ,
correct working (A1)
eg ,
area of A1 N3
[6 marks]
Examiners report
Question
A school café sells three flavours of smoothies: mango (), kiwi fruit () and banana ().
85 students were surveyed about which of these three flavours they like.
35 students liked mango, 37 liked banana, and 26 liked kiwi fruit
2 liked all three flavours
20 liked both mango and banana
14 liked mango and kiwi fruit
3 liked banana and kiwi fruit
Using the given information, complete the following Venn diagram.
Find the number of surveyed students who did not like any of the three flavours.
A student is chosen at random from the surveyed students.
Find the probability that this student likes kiwi fruit smoothies given that they like mango smoothies.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)(A1) (C2)
Note: Award (A1) for 18, 12 and 1 in correct place on Venn diagram, (A1) for 3, 16 and 11 in correct place on Venn diagram.
[2 marks]
85 − (3 + 16 + 11 + 18 + 12 + 1 + 2) (M1)
Note: Award (M1) for subtracting the sum of their values from 85.
22 (A1)(ft) (C2)
Note: Follow through from their Venn diagram in part (a).
If any numbers that are being subtracted are negative award (M1)(A0).
[2 marks]
(A1)(ft)(A1)(ft) (C2)
Note: Award (A1) for correct numerator; (A1) for correct denominator. Follow through from their Venn diagram.
[2 marks]
Examiners report
Question
The diagram shows a circular horizontal board divided into six equal sectors. The sectors are labelled white (W), yellow (Y) and blue (B).
A pointer is pinned to the centre of the board. The pointer is to be spun and when it stops the colour of the sector on which the pointer stops is recorded. The pointer is equally likely to stop on any of the six sectors.
Eva will spin the pointer twice. The following tree diagram shows all the possible outcomes.
Find the probability that both spins are yellow.
Find the probability that at least one of the spins is yellow.
Write down the probability that the second spin is yellow, given that the first spin is blue.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
OR (M1)
Note: Award (M1) for multiplying correct probabilities.
(0.111, 0.111111…, 11.1%) (A1) (C2)
[2 marks]
(M1)(M1)
Note: Award (M1) for and or equivalent, and (M1) for and adding only the three correct probabilities.
OR
(M1)(M1)
Note: Award (M1) for seen and (M1) for subtracting from 1. This may be shown in a tree diagram with “yellow” and “not yellow” branches.
(0.556, 0.555555…, 55.6%) (A1)(ft) (C3)
Note: Follow through marks may be awarded if their answer to part (a) is used in a correct calculation.
[3 marks]
(0.333, 0.333333…, 33.3%) (A1) (C1)
[1 mark]
Examiners report
Question
A cylinder with radius and height is shown in the following diagram.
The sum of and for this cylinder is 12 cm.
Write down an equation for the area, , of the curved surface in terms of .
Find .
Find the value of when the area of the curved surface is maximized.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
OR (A1)(M1) (C2)
Note: Award (A1) for or seen. Award (M1) for correctly substituting into curved surface area of a cylinder. Accept OR .
[2 marks]
(A1)(ft)(A1)(ft) (C2)
Note: Award (A1)(ft) for and (A1)(ft) for . Follow through from part (a). Award at most (A1)(ft)(A0) if additional terms are seen.
[2 marks]
(M1)
Note: Award (M1) for setting their part (b) equal to zero.
6 (cm) (A1)(ft) (C2)
Note: Follow through from part (b).
[2 marks]
Examiners report
Question
A potter sells vases per month.
His monthly profit in Australian dollars (AUD) can be modelled by
Find the value of if no vases are sold.
Differentiate .
Hence, find the number of vases that will maximize the profit.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
−120 (AUD) (A1) (C1)
[1 mark]
(A1)(A1) (C2)
Note: Award (A1) for each correct term. Award at most (A1)(A0) for extra terms seen.
[2 marks]
(M1)
Note: Award (M1) for equating their derivative to zero.
OR
sketch of their derivative (approximately correct shape) with -intercept seen (M1)
(A1)(ft)
Note: Award (C2) for seen without working.
23 (A1)(ft) (C3)
Note: Follow through from part (b).
[3 marks]
Examiners report
Question
A small cuboid box has a rectangular base of length cm and width cm, where . The height is cm, where .
The sum of the length, width and height is cm.
The volume of the box is cm3.
Write down an expression for in terms of .
Find an expression for in terms of .
Find .
Find the value of for which is a maximum.
Justify your answer.
Find the maximum volume.
Markscheme
A1 N1
[1 mark]
correct substitution into volume formula (A1)
eg
A1 N2
Note: Award A0 for unfinished answers such as .
[2 marks]
A1A1 N2
Note: Award A1 for and A1 for .
[2 marks]
valid approach to find maximum (M1)
eg
correct working (A1)
eg
A2 N2
Note: Award A1 for and .
[4 marks]
valid approach to explain that is maximum when (M1)
eg attempt to find , sign chart (must be labelled )
correct value/s A1
eg , where and where
correct reasoning R1
eg , is positive for and negative for
Note: Do not award R1 unless A1 has been awarded.
is maximum when AG N0
[3 marks]
correct substitution into their expression for volume A1
eg ,
(cm3) A1 N1
[2 marks]
Examiners report
Question
Let , for . The point lies on the graph of .
Let . The point lies on the graph of and is the reflection of point in the line .
The line is tangent to the graph of at .
Write down the coordinates of .
Given that , find the equation of in terms of , and .
The line is tangent to the graph of at and has equation .
The line passes through the point .
The gradient of the normal to at is .
Find the equation of in terms of .
Markscheme
(accept ) A2 N2
[2 marks]
Note: There are many approaches to this part, and the steps may be done in any order. Please check working and award marks in line with the markscheme, noting that candidates may work with the equation of the line before finding .
FINDING
valid attempt to find an expression for in terms of (M1)
(A1)
FINDING THE EQUATION OF
EITHER
attempt to substitute tangent gradient and coordinates into equation of straight line (M1)
eg
correct equation in terms of and (A1)
eg
OR
attempt to substitute tangent gradient and coordinates to find
eg
(A1)
THEN (must be in terms of both and )
A1 N3
Note: Award A0 for final answers in the form
[5 marks]
Note: There are many approaches to this part, and the steps may be done in any order. Please check working and award marks in line with the markscheme, noting that candidates may find in terms of before finding a value for .
FINDING
valid approach to find the gradient of the tangent (M1)
eg
correct application of log rule (seen anywhere) (A1)
eg
correct equation (seen anywhere) A1
eg
FINDING
correct substitution of into equation (A1)
eg
(seen anywhere) A1
FINDING
correct substitution of their and into their (A1)
eg
A1 N2
Note: Award A0 for final answers in the form .
[7 marks]
Examiners report
Question
Let . The following diagram shows part of the graph of .
The shaded region is enclosed by the graph of , the -axis and the -axis.
The graph of intersects the -axis at the point .
Find the value of .
Find the volume of the solid formed when the shaded region is revolved about the -axis.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
recognize (M1)
eg
(accept , ) A1 N2
[2 marks]
attempt to substitute either their limits or the function into volume formula (must involve ) (M1)
eg
correct integration of each term A1 A1
eg
substituting limits into their integrated function and subtracting (in any order) (M1)
eg
Note: Award M0 if candidate has substituted into or .
volume A1 N2
[5 marks]
Examiners report
Question
The graph of a function passes through the point .
Given that , find .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
evidence of integration (M1)
eg
correct integration (accept missing ) (A1)
eg
substituting initial condition into their integrated expression (must have ) M1
eg
Note: Award M0 if candidate has substituted into or .
correct application of rule (seen anywhere) (A1)
eg
correct application of rule (seen anywhere) (A1)
eg
correct working (A1)
eg
A1 N4
[7 marks]
Examiners report
Question
In this question, all lengths are in metres and time is in seconds.
Consider two particles, and , which start to move at the same time.
Particle moves in a straight line such that its displacement from a fixed-point is given by , for .
Find an expression for the velocity of at time .
Particle also moves in a straight line. The position of is given by .
The speed of is greater than the speed of when .
Find the value of .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
recognizing velocity is derivative of displacement (M1)
eg
velocity A1 N2
[2 marks]
valid approach to find speed of (M1)
eg , velocity
correct speed (A1)
eg
recognizing relationship between speed and velocity (may be seen in inequality/equation) R1
eg , speed = | velocity | , graph of speed , speed velocity
correct inequality or equation that compares speed or velocity (accept any variable for ) A1
eg
(seconds) (accept , do not accept ) A1 N2
Note: Do not award the last two A1 marks without the R1.
[5 marks]
Examiners report
Question
The following diagram shows part of the graph of , for .
Let be any point on the graph of . Line is the tangent to the graph of at .
Line intersects the -axis at point and the -axis at point B.
Find in terms of and .
Show that the equation of is .
Find the area of triangle in terms of .
The graph of is translated by to give the graph of .
In the following diagram:
- point lies on the graph of
- points , and lie on the vertical asymptote of
- points and lie on the horizontal asymptote of
- point lies on the -axis such that is parallel to .
Line is the tangent to the graph of at , and passes through and .
Given that triangle and rectangle have equal areas, find the gradient of in terms of .
Markscheme
(A1)
A1 N2
[2 marks]
attempt to use point and gradient to find equation of M1
eg
correct working leading to answer A1
eg
AG N0
[2 marks]
METHOD 1 – area of a triangle
recognizing at (M1)
correct working to find -coordinate of null (A1)
eg
-coordinate of null at (may be seen in area formula) A1
correct substitution to find area of triangle (A1)
eg
area of triangle A1 N3
METHOD 2 – integration
recognizing to integrate between and (M1)
eg
correct integration of both terms A1
eg
substituting limits into their integrated function and subtracting (in either order) (M1)
eg
correct working (A1)
eg
area of triangle A1 N3
[5 marks]
Note: In this question, the second M mark may be awarded independently of the other marks, so it is possible to award (M0)(A0)M1(A0)(A0)A0.
recognizing use of transformation (M1)
eg area of triangle = area of triangle , gradient of gradient of , one correct shift
correct working (A1)
eg area of triangle
gradient of area of rectangle
valid approach (M1)
eg
correct working (A1)
eg
correct expression for gradient (in terms of ) (A1)
eg
gradient of is A1 N3
[6 marks]
Examiners report
Question
Consider the graph of the function .
The equation of the tangent to the graph of at is .
Write down .
Write down the gradient of this tangent.
Find the value of .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure. It appeared in a paper that permitted the use of a calculator, and so might not be suitable for all forms of practice.
(A1)(A1)(A1) (C3)
Note: Award (A1) for , (A1) for , and (A1) for or .
Award at most (A1)(A1)(A0) if additional terms are seen.
[3 marks]
(A1) (C1)
[1 mark]
(M1)
Note: Award (M1) for equating their gradient from part (b) to their substituted derivative from part (a).
(A1)(ft) (C2)
Note: Follow through from parts (a) and (b).
[2 marks]
Examiners report
Question
Consider the functions and where .
The graphs of and have a common tangent at .
Find .
Show that .
Hence, show that .
Markscheme
A1
[1 mark]
OR (may be seen anywhere) A1
Note: The derivative of must be explicitly seen, either in terms of or .
recognizing (M1)
OR A1
Note: The final A1 is dependent on one of the previous marks being awarded.
AG
[3 marks]
(M1)
correct equation in
EITHER
A1
A1
OR
A1
A1
THEN
AG
[3 marks]
Examiners report
Question
Consider the function defined by for .
The following diagram shows part of the graph of which crosses the -axis at point , with coordinates . The line is the tangent to the graph of at the point .
Find the exact value of .
Given that the gradient of is , find the -coordinate of .
Markscheme
(M1)
OR (A1)
A1
[3 marks]
attempt to differentiate (must include and/or ) (M1)
A1
setting their derivative M1
OR (or equivalent) A1
valid attempt to solve their quadratic (M1)
A1
Note: Award A0 if the candidate’s final answer includes additional solutions (such as ).
[6 marks]
Examiners report
Question
Let for .
Consider the function defined by for and its graph .
Show that .
The graph of has a horizontal tangent at point . Find the coordinates of .
Given that , show that is a local maximum point.
Solve for .
Sketch the graph of , showing clearly the value of the -intercept and the approximate position of point .
Markscheme
attempt to use quotient or product rule (M1)
OR A1
correct working A1
OR cancelling OR
AG
[3 marks]
(M1)
(A1)
A1
substitution of their to find (M1)
A1
[5 marks]
(M1)
A1
which is negative R1
hence is a local maximum AG
Note: The R1 is dependent on the previous A1 being awarded.
[3 marks]
(A1)
A1
[2 marks]
A1A1A1
Note: Award A1 for one -intercept only, located at
A1 for local maximum, , in approximately correct position
A1 for curve approaching -axis as (including change in concavity).
[3 marks]
Examiners report
Question
Consider the function defined by , for .
The following diagram shows the graph of .
The graph of touches the -axis at points and , as shown. The shaded region is enclosed by the graph of and the -axis, between the points and .
The right cone in the following diagram has a total surface area of , equal to the shaded area in the previous diagram.
The cone has a base radius of , height , and slant height .
Find the -coordinates of and .
Show that the area of the shaded region is .
Find the value of .
Hence, find the volume of the cone.
Markscheme
(or setting their ) (M1)
(or )
A1A1
[3 marks]
attempt to integrate (M1)
A1A1
substitute their limits into their integrated expression and subtract (M1)
A1
area AG
[5 marks]
attempt to substitute into formula for surface area (including base) (M1)
(A1)
A1
[3 marks]
valid attempt to find the height of the cone (M1)
e.g.
(A1)
attempt to use with their values substituted M1
A1
[4 marks]
Examiners report
Question
Particle A travels in a straight line such that its displacement, metres, from a fixed origin after seconds is given by , for , as shown in the following diagram.
Particle A starts at the origin and passes through the origin again when .
Particle A changes direction when .
The total distance travelled by particle A is given by .
Find the value of .
Find the value of .
Find the displacement of particle A from the origin when .
Find the distance of particle A from the origin when .
Find the value of .
A second particle, particle B, travels along the same straight line such that its velocity is given by , for .
When , the distance travelled by particle B is equal to .
Find the value of .
Markscheme
setting (M1)
(accept ) A1
Note: Award A0 if the candidate’s final answer includes additional solutions (such as ).
[2 marks]
recognition that when particle changes direction OR local maximum on graph of OR vertex of parabola (M1)
(accept ) A1
[2 marks]
substituting their value of into OR integrating from to (M1)
A1
[2 marks]
OR OR integrating from to (M1)
A1
[2 marks]
forward backward OR OR (M1)
A1
[2 marks]
METHOD 1
graphical method with triangles on graph M1
(A1)
(A1)
A1
METHOD 2
recognition that distance M1
(A1)
(A1)
A1
[4 marks]
Examiners report
Question
Given that and when , find in terms of .
Markscheme
METHOD 1
recognition that (M1)
(A1)
substitute both and values into their integrated expression including (M1)
A1
METHOD 2
(M1)(A1)
A1
A1
[4 marks]
Examiners report
Question
The function is defined for all . The line with equation is the tangent to the graph of at .
The function is defined for all where and .
Write down the value of .
Find .
Find .
Hence find the equation of the tangent to the graph of at .
Markscheme
A1
[1 mark]
A1
[1 mark]
(M1)
A1
[2 marks]
attempt to use chain rule to find (M1)
OR
A1
OR A1
[3 marks]
Examiners report
Question
A particle moves along the -axis. The velocity of is at time seconds, where for . When is at the origin .
Find the value of when reaches its maximum velocity.
Show that the distance of from at this time is metres.
Sketch a graph of against , clearly showing any points of intersection with the axes.
Find the total distance travelled by .
Markscheme
valid approach to find turning point (, average of roots) (M1)
OR OR
(s) A1
[2 marks]
attempt to integrate (M1)
A1A1
Note: Award A1 for , A1 for .
attempt to substitute their into their solution for the integral (M1)
distance
(or equivalent) A1
(m) AG
[5 marks]
valid approach to solve (may be seen in part (a)) (M1)
OR
correct - intercept on the graph at A1
Note: The following two A marks may only be awarded if the shape is a concave down parabola. These two marks are independent of each other and the (M1).
correct domain from to starting at A1
Note: The must be clearly indicated.
vertex in approximately correct place for and A1
[4 marks]
recognising to integrate between and , or and OR (M1)
A1
A1
valid approach to sum the two areas (seen anywhere) (M1)
OR
total distance travelled (m) A1
[5 marks]
Examiners report
Question
Consider a function with domain . The following diagram shows the graph of , the derivative of .
The graph of , the derivative of , has -intercepts at and . There are local maximum points at and and a local minimum point at .
Find all the values of where the graph of is increasing. Justify your answer.
Find the value of where the graph of has a local maximum.
Find the value of where the graph of has a local minimum. Justify your answer.
Find the values of where the graph of has points of inflexion. Justify your answer.
The total area of the region enclosed by the graph of , the derivative of , and the -axis is .
Given that , find the value of .
Markscheme
Special note: In this question if candidates use the word 'gradient' in their reasoning. e.g. gradient is positive, it must be clear whether this is the gradient of or the gradient of to earn the R mark.
increases when A1
increases when OR is above the -axis R1
Note: Do not award A0R1.
[2 marks]
Special note: In this question if candidates use the word 'gradient' in their reasoning. e.g. gradient is positive, it must be clear whether this is the gradient of or the gradient of to earn the R mark.
A1
[1 mark]
Special note: In this question if candidates use the word 'gradient' in their reasoning. e.g. gradient is positive, it must be clear whether this is the gradient of or the gradient of to earn the R mark.
is minimum when A1
because when and when
(may be seen in a sign diagram clearly labelled as )
OR because changes from negative to positive at
OR and slope of is positive at R1
Note: Do not award A0 R1
[2 marks]
Special note: In this question if candidates use the word 'gradient' in their reasoning. e.g. gradient is positive, it must be clear whether this is the gradient of or the gradient of to earn the R mark.
has points of inflexion when and A2
has turning points at and
OR
and and changes from increasing to decreasing or vice versa at each of these -values (may be seen in a sign diagram clearly labelled as and ) R1
Note: Award A0 if any incorrect answers are given. Do not award A0R1
[3 marks]
Special note: In this question if candidates use the word 'gradient' in their reasoning. e.g. gradient is positive, it must be clear whether this is the gradient of or the gradient of to earn the R mark.
recognizing area from to (seen anywhere) M1
recognizing to negate integral for area below -axis (M1)
OR
(for any integral) (M1)
OR (A1)
(A1)
A1
[6 marks]
Examiners report
Question
The expression can be written as . Write down the value of .
Hence, find the value of .
Markscheme
A1
[1 mark]
A1A1
substituting limits into their integrated function and subtracting (M1)
OR
A1
[4 marks]
Examiners report
Many candidates could give the value of p correctly. However, many did struggle with the integration, including substituting limits into the integrand, without integrating at all. An incorrect value of p often resulted in arithmetic of greater complexity.
Question
Consider the curve with equation , where and .
The tangent to the curve at the point where is parallel to the line .
Find the value of .
Markscheme
evidence of using product rule (M1)
A1
correct working for one of (seen anywhere) A1
at
OR
slope of tangent is
their at equals the slope of (seen anywhere) (M1)
A1
[5 marks]
Examiners report
The product rule was well recognised and used with 𝑥=1 properly substituted into this expression. Although the majority of the candidates tried equating the derivative to the slope of the tangent line, the slope of the tangent line was not correctly identified; many candidates incorrectly substituted 𝑥=1 into the tangent equation, thus finding the y-coordinate instead of the slope.
Question
The following diagram shows part of the graph of a quadratic function .
The graph of has its vertex at , and it passes through point as shown.
The function can be written in the form .
The line is tangent to the graph of at .
Now consider another function . The derivative of is given by , where .
Write down the equation of the axis of symmetry.
Write down the values of and .
Point has coordinates . Find the value of .
Find the equation of .
Find the values of for which is an increasing function.
Find the values of for which the graph of is concave-up.
Markscheme
A1
Note: Must be an equation in the form “ ”. Do not accept or .
[1 mark]
(accept ) A1A1
[2 marks]
attempt to substitute coordinates of (M1)
A1
[2 marks]
recognize need to find derivative of (M1)
or A1
(may be seen as gradient in their equation) (A1)
or A1
Note: Award A0 for .
[4 marks]
METHOD 1
Recognizing that for to be increasing, , or (M1)
The vertex must be above the -axis, (R1)
A1
METHOD 2
attempting to find discriminant of (M1)
recognizing discriminant must be negative (R1)
OR
A1
[3 marks]
recognizing that for to be concave up, (M1)
when (R1)
A1
[3 marks]
Examiners report
In parts (a) and (b) of this question, a majority of candidates recognized the connection between the coordinates of the vertex and the axis of symmetry and the values of and , and most candidates were able to successfully substitute the coordinates of point Q to find the value of . In part (c), the candidates who recognized the need to use the derivative to find the gradient of the tangent were generally successful in finding the equation of the line, although many did not give their equation in the proper form in terms of and , and instead wrote , thus losing the final mark. Parts (d) and (e) were much more challenging for candidates. Although a good number of candidates recognized that in part (d), and in part (e), very few were able to proceed beyond this point and find the correct inequalities for their final answers.
Question
Consider the functions , for , and for .
The following diagram shows the graphs of and .
The graphs of and intersect at points and . The coordinates of are .
In the following diagram, the shaded region is enclosed by the graph of , the graph of , the -axis, and the line , where .
The area of the shaded region can be written as , where .
Find the coordinates of .
Find the value of and the value of .
Markscheme
(M1)
OR (A1)
valid attempt to solve their quadratic (M1)
OR OR
(may be seen in answer) A1
(accept ) A1
[5 marks]
recognizing two correct regions from to and from to (R1)
triangle OR OR
area of triangle is OR OR (A1)
correct integration (A1)(A1)
Note: Award A1 for and A1 for .
Note: The first three A marks may be awarded independently of the R mark.
substitution of their limits (for ) into their integrated function (in terms of ) (M1)
A1
adding their two areas (in terms of ) and equating to (M1)
equating their non-log terms to (equation must be in terms of ) (M1)
A1
A1
[10 marks]
Examiners report
Nearly all candidates knew to set up an equation with in order to find the intersection of the two graphs, and most were able to solve the resulting quadratic equation. Candidates were not as successful in part (b), however. While some candidates recognized that there were two regions to be added together, very few were able to determine the correct boundaries of these regions, with many candidates integrating one or both functions from to . While a good number of candidates were able to correctly integrate the function(s), without the correct bounds the values of and were unattainable.
Question
A function, , has its derivative given by , where . The following diagram shows part of the graph of .
The graph of has an axis of symmetry .
The vertex of the graph of lies on the -axis.
The graph of has a point of inflexion at .
Find the value of .
Write down the value of the discriminant of .
Hence or otherwise, find the value of .
Find the value of the gradient of the graph of at .
Sketch the graph of , the second derivative of . Indicate clearly the -intercept and the -intercept.
Write down the value of .
Find the values of for which the graph of is concave-down. Justify your answer.
Markscheme
EITHER
attempt to use (M1)
OR
attempt to complete the square (M1)
OR
attempt to differentiate and equate to (M1)
THEN
A1
[2 marks]
discriminant A1
[1 mark]
EITHER
attempt to substitute into (M1)
A1
OR
(M1)
A1
THEN
A1
[3 marks]
A1
attempt to find (M1)
gradient A1
[3 marks]
A1A1
Note: Award A1 for line with positive gradient, A1 for correct intercepts.
[2 marks]
A1
[1 mark]
A1
(for ) OR the is below the -axis (for )
OR (sign diagram must be labelled ) R1
[2 marks]
Examiners report
Candidates did score well on this question. As always, candidates are encouraged to read the questions carefully for key words such as 'value' as opposed to 'expression'. So, if asked for the value of the discriminant, their answer should be a number and not an expression found from . As such the value of the discriminant in (b)(i) was often seen in (b)(ii). Please ask students to use a straight edge when sketching a straight line! Overall, the reasoning mark for determining where the graph of f is concave-down, was an improvement on previous years. Sign diagrams were typically well labelled, and the description contained clarity regarding which function was being referred to.
Question
Consider .
Expand and simplify in ascending powers of .
By using a suitable substitution for , show that .
Show that , where is a positive real constant.
It is given that , where . Find the value of .
Markscheme
EITHER
attempt to use binomial expansion (M1)
OR
(M1)
THEN
A1
[2 marks]
(A1)
So,
A1
attempt to substitute any double angle rule for into (M1)
A1
AG
Note: Allow working RHS to LHS.
[4 marks]
recognizing to integrate (M1)
EITHER
applies integration by inspection (M1)
A1
A1
OR
(M1)
A1
OR A1
THEN
AG
[4 marks]
EITHER
M1
OR (M1)
OR
M1
(M1)
THEN
(A1)
(A1)
A1
[5 marks]
Examiners report
Many candidates successfully expanded the binomial, with the most common error being to omit the negative sign with a. The connection between (a)(i) and (ii) was often noted but not fully utilised with candidates embarking on unnecessary complex algebraic expansions of expressions involving double angle rules. Candidates often struggled to apply inspection or substitution when integrating. As a 'show that' question, b(i) provided a useful result to be utilised in (ii). So even without successfully completing (i) candidates could apply it in part (ii). Not many managed to do so.
Question
The derivative of a function is given by .
Given that , find the value of .
Markscheme
* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
correct working (A1)
eg −5 + (8 − 1)(3)
u8 = 16 A1 N2
[2 marks]
METHOD 1
(A1)
attempts to integrate (M1)
A1
uses to obtain and so M1
substitutes into their expression for (M1)
so A1
METHOD 2
(A1)
attempts to integrate both sides (M1)
A1
M1
uses to find their value of (M1)
so A1
[6 marks]
Examiners report
Question
The following diagram shows the graph of , and rectangle . The rectangle has a vertex at the origin , a vertex on the -axis at the point , a vertex on the -axis at the point and a vertex at point on the graph.
Let represent the perimeter of rectangle .
Let represent the area of rectangle .
Show that .
Find the dimensions of rectangle that has maximum perimeter and determine the value of the maximum perimeter.
Find an expression for in terms of .
Find the dimensions of rectangle that has maximum area.
Determine the maximum area of rectangle .
Markscheme
* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
(A1)
A1
so AG
[2 marks]
METHOD 1
EITHER
uses the axis of symmetry of a quadratic (M1)
OR
forms (M1)
THEN
A1
substitutes their value of into (M1)
A1
so the dimensions of rectangle of maximum perimeter are by
EITHER
substitutes their value of into (M1)
OR
substitutes their values of and into (M1)
A1
so the maximum perimeter is
METHOD 2
attempts to complete the square M1
A1
A1
substitutes their value of into (M1)
A1
so the dimensions of rectangle of maximum perimeter are by
A1
so the maximum perimeter is
[6 marks]
substitutes into (M1)
A1
[2 marks]
A1
attempts to solve their for (M1)
A1
substitutes their (positive) value of into (M1)
A1
[5 marks]
A1
[1 mark]
Examiners report
Question
The following diagram shows a ball attached to the end of a spring, which is suspended from a ceiling.
The height, metres, of the ball above the ground at time seconds after being released can be modelled by the function where .
Find the height of the ball above the ground when it is released.
Find the minimum height of the ball above the ground.
Show that the ball takes seconds to return to its initial height above the ground for the first time.
For the first 2 seconds of its motion, determine the amount of time that the ball is less than metres above the ground.
Find the rate of change of the ball’s height above the ground when . Give your answer in the form where and .
Markscheme
* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
attempts to find (M1)
(m) (above the ground) A1
[2 marks]
EITHER
uses the minimum value of which is M1
(m)
OR
the amplitude of motion is (m) and the mean position is (m) M1
OR
finds , attempts to solve for and determines that the minimum height above the ground occurs at M1
(m)
THEN
(m) (above the ground) A1
[2 marks]
EITHER
the ball is released from its maximum height and returns there a period later R1
the period is A1
OR
attempts to solve for M1
A1
THEN
so it takes seconds for the ball to return to its initial position for the first time AG
[2 marks]
(M1)
A1
(A1)
Note: Accept extra correct positive solutions for .
A1
Note: Do not award A1 if solutions outside are also stated.
the ball is less than metres above the ground for (s)
(s) A1
[5 marks]
EITHER
attempts to find (M1)
OR
recognizes that is required (M1)
THEN
A1
attempts to evaluate their (M1)
A1
Note: Accept equivalent correct answer forms where . For example, .
[4 marks]
Examiners report
Question
Let . Given that , find .
Markscheme
attempt to integrate (M1)
(A1)
EITHER
A1
OR
A1
THEN
correct substitution into their integrated function (must have C) (M1)
A1
[5 marks]
Examiners report
Question
Let .
The graph of has horizontal tangents at the points where = and = , < .
Find .
Find the value of and the value of .
Sketch the graph of .
Hence explain why the graph of has a local maximum point at .
Find .
Hence, use your answer to part (d)(i) to show that the graph of has a local minimum point at .
The normal to the graph of at and the tangent to the graph of at intersect at the point (, ) .
Find the value of and the value of .
Markscheme
(M1)A1
[2 marks]
correct reasoning that (seen anywhere) (M1)
valid approach to solve quadratic M1
, quadratic formula
correct values for
3, −5
correct values for and
= −5 and = 3 A1
[3 marks]
A1
[1 mark]
first derivative changes from positive to negative at A1
so local maximum at AG
[1 mark]
A1
substituting their into their second derivative (M1)
(A1)
[3 marks]
is positive so graph is concave up R1
so local minimum at AG
[1 mark]
normal to at is = −5 (seen anywhere) (A1)
attempt to find -coordinate at their value of (M1)
−10 (A1)
tangent at has equation = −10 (seen anywhere) A1
intersection at (−5, −10)
= −5 and = −10 A1
[5 marks]
Examiners report
Question
Let where , .
The graph of has exactly one maximum point P.
The second derivative of is given by . The graph of has exactly one point of inflexion Q.
Show that .
Find the x-coordinate of P.
Show that the x-coordinate of Q is .
The region R is enclosed by the graph of , the x-axis, and the vertical lines through the maximum point P and the point of inflexion Q.
Given that the area of R is 3, find the value of .
Markscheme
attempt to use quotient rule (M1)
correct substitution into quotient rule
(or equivalent) A1
, A1
AG
[3 marks]
M1
(A1)
A1
[3 marks]
M1
A1
A1
so the point of inflexion occurs at AG
[3 marks]
attempt to integrate (M1)
(A1)
EITHER
= A1
so A1
OR
A1
so A1
THEN
A1
setting their expression for area equal to 3 M1
A1
[7 marks]
Examiners report
Question
Let and .
The graphs of and intersect at and , where .
Find the value of and of .
Hence, find the area of the region enclosed by the graphs of and .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid attempt to find the intersection (M1)
eg, sketch, one correct answer
A1A1 N3
[3 marks]
attempt to set up an integral involving subtraction (in any order) (M1)
eg
0.537667
A2 N3
[3 marks]
Examiners report
Question
All lengths in this question are in metres.
Let , for . Mark uses as a model to create a barrel. The region enclosed by the graph of , the -axis, the line and the line is rotated 360° about the -axis. This is shown in the following diagram.

Use the model to find the volume of the barrel.
The empty barrel is being filled with water. The volume of water in the barrel after minutes is given by . How long will it take for the barrel to be half-full?
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to substitute correct limits or the function into the formula involving
eg
0.601091
volume A2 N3
[3 marks]
attempt to equate half their volume to (M1)
eg, graph
4.71104
4.71 (minutes) A2 N3
[3 marks]
Examiners report
Question
A particle P starts from a point A and moves along a horizontal straight line. Its velocity after seconds is given by
The following diagram shows the graph of .

P is at rest when and .
When , the acceleration of P is zero.
Find the initial velocity of .
Find the value of .
(i) Find the value of .
(ii) Hence, find the speed of P when .
(i) Find the total distance travelled by P between and .
(ii) Hence or otherwise, find the displacement of P from A when .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid attempt to substitute into the correct function (M1)
eg
2 A1 N2
[2 marks]
recognizing when P is at rest (M1)
5.21834
A1 N2
[2 marks]
(i) recognizing that (M1)
eg, minimum on graph
1.95343
A1 N2
(ii) valid approach to find their minimum (M1)
eg, reference to min on graph
1.75879
speed A1 N2
[4 marks]
(i) substitution of correct into distance formula, (A1)
eg
4.45368
distance A1 N2
(ii) displacement from to (seen anywhere) (A1)
eg
displacement from to (A1)
eg
valid approach to find displacement for M1
eg
displacement A1 N2
[6 marks]
Examiners report
Question
A group of 66 people went on holiday to Hawaii. During their stay, three trips were arranged: a boat trip (), a coach trip () and a helicopter trip ().
From this group of people:
| 3 | went on all three trips; |
| 16 | went on the coach trip only; |
| 13 | went on the boat trip only; |
| 5 | went on the helicopter trip only; |
| x | went on the coach trip and the helicopter trip but not the boat trip; |
| 2x | went on the boat trip and the helicopter trip but not the coach trip; |
| 4x | went on the boat trip and the coach trip but not the helicopter trip; |
| 8 | did not go on any of the trips. |
One person in the group is selected at random.
Draw a Venn diagram to represent the given information, using sets labelled , and .
Show that .
Write down the value of .
Find the probability that this person
(i) went on at most one trip;
(ii) went on the coach trip, given that this person also went on both the helicopter trip and the boat trip.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A5)
Notes: Award (A1) for rectangle and three labelled intersecting circles (U need not be seen),
(A1) for 3 in the correct region,
(A1) for 8 in the correct region,
(A1) for 5, 13 and 16 in the correct regions,
(A1) for , and in the correct regions.
[5 marks]
(M1)
Note: Award (M1) for either a completely correct equation or adding all the terms from their diagram in part (a) and equating to 66.
Award (M0)(A0) if their equation has no .
OR (A1)
Note: Award (A1) for adding their like terms correctly, but only when the solution to their equation is equal to 3 and is consistent with their original equation.
(AG)
Note: The conclusion must be seen for the (A1) to be awarded.
[2 marks]
15 (A1)(ft)
Note: Follow through from part (a). The answer must be an integer.
[1 mark]
(i) (A1)(ft)(A1)(G2)
Note: Award (A1)(ft) for numerator, (A1) for denominator. Follow through from their Venn diagram.
(ii) (A1)(A1)(ft)(G2)
Note: Award (A1) for numerator, (A1)(ft) for denominator. Follow through from their Venn diagram.
[4 marks]
Examiners report
Question
A water container is made in the shape of a cylinder with internal height cm and internal base radius cm.

The water container has no top. The inner surfaces of the container are to be coated with a water-resistant material.
The volume of the water container is .
The water container is designed so that the area to be coated is minimized.
One can of water-resistant material coats a surface area of .
Write down a formula for , the surface area to be coated.
Express this volume in .
Write down, in terms of and , an equation for the volume of this water container.
Show that .
Find .
Using your answer to part (e), find the value of which minimizes .
Find the value of this minimum area.
Find the least number of cans of water-resistant material that will coat the area in part (g).
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)(A1)
Note: Award (A1) for either OR seen. Award (A1) for two correct terms added together.
[2 marks]
(A1)
Notes: Units not required.
[1 mark]
(A1)(ft)
Notes: Award (A1)(ft) for equating to their part (b).
Do not accept unless is explicitly defined as their part (b).
[1 mark]
(A1)(ft)(M1)
Note: Award (A1)(ft) for their seen.
Award (M1) for correctly substituting only into a correct part (a).
Award (A1)(ft)(M1) for rearranging part (c) to and substituting for in expression for .
(AG)
Notes: The conclusion, , must be consistent with their working seen for the (A1) to be awarded.
Accept as equivalent to .
[2 marks]
(A1)(A1)(A1)
Note: Award (A1) for , (A1) for or , (A1) for .
[3 marks]
(M1)
Note: Award (M1) for equating their part (e) to zero.
OR (M1)
Note: Award (M1) for isolating .
OR
sketch of derivative function (M1)
with its zero indicated (M1)
(A1)(ft)(G2)
[3 marks]
(M1)
Note: Award (M1) for correct substitution of their part (f) into the given equation.
(A1)(ft)(G2)
[2 marks]
(M1)
Note: Award (M1) for dividing their part (g) by 2000.
(A1)(ft)
Notes: Follow through from part (g).
14 (cans) (A1)(ft)(G3)
Notes: Final (A1) awarded for rounding up their to the next integer.
[3 marks]
Examiners report
Question
Let . Find the term in in the expansion of the derivative, .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
derivative of A2
recognizing need to find term in (seen anywhere) R1
eg
valid approach to find the terms in (M1)
eg, Pascal’s triangle to 6th row
identifying correct term (may be indicated in expansion) (A1)
eg
correct working (may be seen in expansion) (A1)
eg
A1 N3
METHOD 2
recognition of need to find in (seen anywhere) R1
valid approach to find the terms in (M1)
eg, Pascal’s triangle to 7th row
identifying correct term (may be indicated in expansion) (A1)
eg6th term,
correct working (may be seen in expansion) (A1)
eg
correct term (A1)
differentiating their term in (M1)
eg
A1 N3
[7 marks]
Examiners report
Question
Note: In this question, distance is in metres and time is in seconds.
A particle moves along a horizontal line starting at a fixed point A. The velocity of the particle, at time , is given by , for . The following diagram shows the graph of

There are -intercepts at and .
Find the maximum distance of the particle from A during the time and justify your answer.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1 (displacement)
recognizing (M1)
consideration of displacement at and (seen anywhere) M1
eg and
Note: Must have both for any further marks.
correct displacement at and (seen anywhere) A1A1
(accept 2.28318), 1.55513
valid reasoning comparing correct displacements R1
eg, more left than right
2.28 (m) A1 N1
Note: Do not award the final A1 without the R1.
METHOD 2 (distance travelled)
recognizing distance (M1)
consideration of distance travelled from to 2 and to 5 (seen anywhere) M1
eg and
Note: Must have both for any further marks
correct distances travelled (seen anywhere) A1A1
2.28318, (accept ), 3.83832
valid reasoning comparing correct distance values R1
eg
2.28 (m) A1 N1
Note: Do not award the final A1 without the R1.
[6 marks]
Examiners report
Question
A particle P moves along a straight line. Its velocity after seconds is given by , for . The following diagram shows the graph of .

Write down the first value of at which P changes direction.
Find the total distance travelled by P, for .
A second particle Q also moves along a straight line. Its velocity, after seconds is given by for . After seconds Q has travelled the same total distance as P.
Find .
Markscheme
A1 N1
[1 mark]
substitution of limits or function into formula or correct sum (A1)
eg
9.64782
distance A1 N2
[2 marks]
correct approach (A1)
eg
correct integration (A1)
eg
equating their expression to the distance travelled by their P (M1)
eg
5.93855
5.94 (seconds) A1 N3
[4 marks]
Examiners report
Question
Let . The following diagram shows part of the graph of .

There are -intercepts at and at . There is a maximum at A where , and a point of inflexion at B where .
Find the value of .
Write down the coordinates of A.
Write down the rate of change of at A.
Find the coordinates of B.
Find the the rate of change of at B.
Let be the region enclosed by the graph of , the -axis, the line and the line . The region is rotated 360° about the -axis. Find the volume of the solid formed.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
evidence of valid approach (M1)
eg
2.73205
A1 N2
[2 marks]
1.87938, 8.11721
A2 N2
[2 marks]
rate of change is 0 (do not accept decimals) A1 N1
[1 marks]
METHOD 1 (using GDC)
valid approach M1
eg, max/min on
sketch of either or , with max/min or root (respectively) (A1)
A1 N1
Substituting their value into (M1)
eg
A1 N1
METHOD 2 (analytical)
A1
setting (M1)
A1 N1
substituting their value into (M1)
eg
A1 N1
[4 marks]
recognizing rate of change is (M1)
eg
rate of change is 6 A1 N2
[3 marks]
attempt to substitute either limits or the function into formula (M1)
involving (accept absence of and/or )
eg
128.890
A2 N3
[3 marks]
Examiners report
Question
Let and , for .
The graph of can be obtained from the graph of by two transformations:
Let , for . The following diagram shows the graph of and the line .

The graph of intersects the graph of at two points. These points have coordinates 0.111 and 3.31 correct to three significant figures.
Write down the value of ;
Write down the value of ;
Write down the value of .
Find .
Hence, find the area of the region enclosed by the graphs of and .
Let be the vertical distance from a point on the graph of to the line . There is a point on the graph of where is a maximum.
Find the coordinates of P, where .
Markscheme
A1 N1
Note: Accept , , and , 2.31 as candidate may have rewritten as equal to .
[1 mark]
A1 N1
Note: Accept , , and , 2.31 as candidate may have rewritten as equal to .
[1 mark]
A1 N1
Note: Accept , , and , 2.31 as candidate may have rewritten as equal to .
[1 mark]
2.72409
2.72 A2 N2
[2 marks]
recognizing area between and equals 2.72 (M1)
eg
recognizing graphs of and are reflections of each other in (M1)
egarea between and equals between and
5.44819
5.45 A1 N3
[??? marks]
valid attempt to find (M1)
egdifference in -coordinates,
correct expression for (A1)
eg
valid approach to find when is a maximum (M1)
egmax on sketch of , attempt to solve
0.973679
A2 N4
substituting their value into (M1)
2.26938
A1 N2
[7 marks]
Examiners report
Question
Violeta plans to grow flowers in a rectangular plot. She places a fence to mark out the perimeter of the plot and uses 200 metres of fence. The length of the plot is metres.

Violeta places the fence so that the area of the plot is maximized.
By selling her flowers, Violeta earns 2 Bulgarian Levs (BGN) per square metre of the plot.
Show that the width of the plot, in metres, is given by .
Write down the area of the plot in terms of .
Find the value of that maximizes the area of the plot.
Show that Violeta earns 5000 BGN from selling the flowers grown on the plot.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(or equivalent) (M1)
OR
(or equivalent) (M1)
Note: Award (M1) for a correct expression leading to (the does not need to be seen). The 200 must be seen for the (M1) to be awarded. Do not accept substituted in the perimeter of the rectangle formula.
(AG)
[1 mark]
OR (or equivalent) (A1)
[1 mark]
ORORgraphical method (M1)
Note: Award (M1) for use of axis of symmetry formula or first derivative equated to zero or a sketch graph.
(A1)(ft)(G2)
Note: Follow through from part (b), provided x is positive and less than 100.
[2 marks]
(M1)(M1)
Note: Award (M1) for substituting their into their formula for area (accept “” for the substituted formula), and (M1) for multiplying by 2. Award at most (M0)(M1) if their calculation does not lead to 5000 (BGN), although the 5000 (BGN) does not need to be seen explicitly.
Substitution of 50 into area formula may be seen in part (c).
5000 (BGN) (AG)
[2 marks]
Examiners report
Question
Consider the function , where is a constant. Part of the graph of is shown below.

It is known that at the point where the tangent to the graph of is horizontal.
There are two other points on the graph of at which the tangent is horizontal.
Write down the -intercept of the graph.
Find .
Show that .
Find .
Write down the -coordinates of these two points;
Write down the intervals where the gradient of the graph of is positive.
Write down the range of .
Write down the number of possible solutions to the equation .
The equation , where , has four solutions. Find the possible values of .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
5 (A1)
Note: Accept an answer of .
[1 mark]
(A1)(A1)
Note: Award (A1) for and (A1) for . Award at most (A1)(A0) if extra terms are seen.
[2 marks]
(M1)(M1)
Note: Award (M1) for substitution of into their derivative, (M1) for equating their derivative, written in terms of , to 0 leading to a correct answer (note, the 8 does not need to be seen).
(AG)
[2 marks]
(M1)
Note: Award (M1) for correct substitution of and into the formula of the function.
21 (A1)(G2)
[2 marks]
(A1)(A1)
Note: Award (A1) for each correct solution. Award at most (A0)(A1)(ft) if answers are given as and or and .
[2 marks]
(A1)(ft)(A1)(ft)
Note: Award (A1)(ft) for , follow through from part (d)(i) provided their value is negative.
Award (A1)(ft) for , follow through only from their 0 from part (d)(i); 2 must be the upper limit.
Accept interval notation.
[2 marks]
(A1)(ft)(A1)
Notes: Award (A1)(ft) for 21 seen in an interval or an inequality, (A1) for “”.
Accept interval notation.
Accept or .
Follow through from their answer to part (c)(ii). Award at most (A1)(ft)(A0) if is seen instead of . Do not award the second (A1) if a (finite) lower limit is seen.
[2 marks]
3 (solutions) (A1)
[1 mark]
or equivalent (A1)(ft)(A1)
Note: Award (A1)(ft) for 5 and 21 seen in an interval or an inequality, (A1) for correct strict inequalities. Follow through from their answers to parts (a) and (c)(ii).
Accept interval notation.
[2 marks]
Examiners report
Question
Let , for . The graph of passes through the point , where .
Find the value of .
The following diagram shows part of the graph of .

The region enclosed by the graph of , the -axis and the lines and is rotated 360° about the -axis. Find the volume of the solid formed.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach (M1)
eg, intersection with
2.32143
(exact), 2.32 A1 N2
[2 marks]
attempt to substitute either their limits or the function into volume formula (must involve , accept reversed limits and absence of and/or , but do not accept any other errors) (M1)
eg
331.989
A2 N3
[3 marks]
Examiners report
Question
Note: In this question, distance is in metres and time is in seconds.
A particle P moves in a straight line for five seconds. Its acceleration at time is given by , for .
When , the velocity of P is .
Write down the values of when .
Hence or otherwise, find all possible values of for which the velocity of P is decreasing.
Find an expression for the velocity of P at time .
Find the total distance travelled by P when its velocity is increasing.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
A1A1 N2
[2 marks]
recognizing that is decreasing when is negative (M1)
eg, sketch of
correct interval A1 N2
eg
[2 marks]
valid approach (do not accept a definite integral) (M1)
eg
correct integration (accept missing ) (A1)(A1)(A1)
substituting , (must have ) (M1)
eg
A1 N6
[6 marks]
recognizing that increases outside the interval found in part (b) (M1)
eg, diagram
one correct substitution into distance formula (A1)
eg
one correct pair (A1)
eg3.13580 and 11.0833, 20.9906 and 35.2097
14.2191 A1 N2
[4 marks]
Examiners report
Question
A company performs an experiment on the efficiency of a liquid that is used to detect a nut allergy.
A group of 60 people took part in the experiment. In this group 26 are allergic to nuts. One person from the group is chosen at random.
A second person is chosen from the group.
When the liquid is added to a person’s blood sample, it is expected to turn blue if the person is allergic to nuts and to turn red if the person is not allergic to nuts.
The company claims that the probability that the test result is correct is 98% for people who are allergic to nuts and 95% for people who are not allergic to nuts.
It is known that 6 in every 1000 adults are allergic to nuts.
This information can be represented in a tree diagram.

An adult, who was not part of the original group of 60, is chosen at random and tested using this liquid.
The liquid is used in an office to identify employees who might be allergic to nuts. The liquid turned blue for 38 employees.
Find the probability that both people chosen are not allergic to nuts.
Copy and complete the tree diagram.
Find the probability that this adult is allergic to nuts and the liquid turns blue.
Find the probability that the liquid turns blue.
Find the probability that the tested adult is allergic to nuts given that the liquid turned blue.
Estimate the number of employees, from this 38, who are allergic to nuts.
Markscheme
(M1)
Note: Award (M1) for their correct product.
(A1)(ft)(G2)
Note: Follow through from part (a).
[2 marks]
(A1)(A1)(A1)
Note: Award (A1) for each correct pair of branches.
[3 marks]
(M1)
Note: Award (M1) for multiplying 0.006 by 0.98.
(A1)(G2)
[2 marks]
(A1)(ft)(M1)
Note: Award (A1)(ft) for their two correct products, (M1) for adding two products.
(A1)(ft)(G3)
Note: Follow through from parts (c) and (d).
[3 marks]
(M1)(M1)
Note: Award (M1) for their correct numerator, (M1) for their correct denominator.
(A1)(ft)(G3)
Note: Follow through from parts (d) and (e).
[3 marks]
(M1)
Note: Award (M1) for multiplying 38 by their answer to part (f).
(A1)(ft)(G2)
Notes: Follow through from part (f). Use of 3 sf result from part (f) results in an answer of 4.03 (4.028).
[2 marks]
Examiners report
Question
A function is given by .
The graph of the function intersects the graph of .
Expand the expression for .
Find .
Draw the graph of for and . Use a scale of 2 cm to represent 1 unit on the -axis and 1 cm to represent 5 units on the -axis.
Write down the coordinates of the point of intersection.
Markscheme
(A1)
Notes: The expansion may be seen in part (b)(ii).
[1 mark]
(A1)(ft)(A1)(ft)(A1)(ft)
Notes: Follow through from part (b)(i). Award (A1)(ft) for each correct term. Award at most (A1)(ft)(A1)(ft)(A0) if extra terms are seen.
[3 marks]
(A1)(A1)(ft)(A1)(ft)(A1)
Notes: Award (A1) for correct scale; axes labelled and drawn with a ruler.
Award (A1)(ft) for their correct -intercepts in approximately correct location.
Award (A1) for correct minimum and maximum points in approximately correct location.
Award (A1) for a smooth continuous curve with approximate correct shape. The curve should be in the given domain.
Follow through from part (a) for the -intercepts.
[4 marks]
(G1)(ft)(G1)(ft)
Notes: Award (G1) for 1.49 and (G1) for 13.9 written as a coordinate pair. Award at most (G0)(G1) if parentheses are missing. Accept and . Follow through from part (b)(i).
[2 marks]
Examiners report
Question
Let f(x) = ln x − 5x , for x > 0 .
Solve f '(x) = f "(x).
Markscheme
METHOD 1 (using GDC)
valid approach (M1)
eg
0.558257
x = 0.558 A1 N2
Note: Do not award A1 if additional answers given.
METHOD 2 (analytical)
attempt to solve their equation f '(x) = f "(x) (do not accept ) (M1)
eg
0.558257
x = 0.558 A1 N2
Note: Do not award A1 if additional answers given.
[2 marks]
Examiners report
Question
Let for 0 ≤ ≤ 1.5. The following diagram shows the graph of .
Find the x-intercept of the graph of .
The region enclosed by the graph of , the y-axis and the x-axis is rotated 360° about the x-axis.
Find the volume of the solid formed.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach (M1)
eg or 0…
1.14472
(exact), 1.14 A1 N2
[2 marks]
attempt to substitute either their limits or the function into formula involving . (M1)
eg
2.49799
volume = 2.50 A2 N3
[3 marks]
Examiners report
Question
Let g(x) = −(x − 1)2 + 5.
Let f(x) = x2. The following diagram shows part of the graph of f.
The graph of g intersects the graph of f at x = −1 and x = 2.
Write down the coordinates of the vertex of the graph of g.
On the grid above, sketch the graph of g for −2 ≤ x ≤ 4.
Find the area of the region enclosed by the graphs of f and g.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(1,5) (exact) A1 N1
[1 mark]
A1A1A1 N3
Note: The shape must be a concave-down parabola.
Only if the shape is correct, award the following for points in circles:
A1 for vertex,
A1 for correct intersection points,
A1 for correct endpoints.
[3 marks]
integrating and subtracting functions (in any order) (M1)
eg
correct substitution of limits or functions (accept missing dx, but do not accept any errors, including extra bits) (A1)
eg
area = 9 (exact) A1 N2
[3 marks]
Examiners report
Question
Let , for x > 0.
The k th maximum point on the graph of f has x-coordinate xk where .
Given that xk + 1 = xk + a, find a.
Hence find the value of n such that .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach to find maxima (M1)
eg one correct value of xk, sketch of f
any two correct consecutive values of xk (A1)(A1)
eg x1 = 1, x2 = 5
a = 4 A1 N3
[4 marks]
recognizing the sequence x1, x2, x3, …, xn is arithmetic (M1)
eg d = 4
correct expression for sum (A1)
eg
valid attempt to solve for n (M1)
eg graph, 2n2 − n − 861 = 0
n = 21 A1 N2
[4 marks]
Examiners report
Question
A particle P moves along a straight line. The velocity v m s−1 of P after t seconds is given by v (t) = 7 cos t − 5t cos t, for 0 ≤ t ≤ 7.
The following diagram shows the graph of v.
Find the initial velocity of P.
Find the maximum speed of P.
Write down the number of times that the acceleration of P is 0 m s−2 .
Find the acceleration of P when it changes direction.
Find the total distance travelled by P.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
initial velocity when t = 0 (M1)
eg v(0)
v = 7 (m s−1) A1 N2
[2 marks]
recognizing maximum speed when is greatest (M1)
eg minimum, maximum, v' = 0
one correct coordinate for minimum (A1)
eg 6.37896, −24.6571
24.7 (ms−1) A1 N2
[3 marks]
recognizing a = v ′ (M1)
eg , correct derivative of first term
identifying when a = 0 (M1)
eg turning points of v, t-intercepts of v ′
3 A1 N3
[3 marks]
recognizing P changes direction when v = 0 (M1)
t = 0.863851 (A1)
−9.24689
a = −9.25 (ms−2) A2 N3
[4 marks]
correct substitution of limits or function into formula (A1)
eg
63.8874
63.9 (metres) A2 N3
[3 marks]
Examiners report
Question
Let , be a periodic function with
The following diagram shows the graph of .
There is a maximum point at A. The minimum value of is −13 .
A ball on a spring is attached to a fixed point O. The ball is then pulled down and released, so that it moves back and forth vertically.
The distance, d centimetres, of the centre of the ball from O at time t seconds, is given by
Find the coordinates of A.
For the graph of , write down the amplitude.
For the graph of , write down the period.
Hence, write in the form .
Find the maximum speed of the ball.
Find the first time when the ball’s speed is changing at a rate of 2 cm s−2.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
−0.394791,13
A(−0.395, 13) A1A1 N2
[2 marks]
13 A1 N1
[1 mark]
, 6.28 A1 N1
[1 mark]
valid approach (M1)
eg recognizing that amplitude is p or shift is r
(accept p = 13, r = 0.395) A1A1 N3
Note: Accept any value of r of the form
[3 marks]
recognizing need for d ′(t) (M1)
eg −12 sin(t) − 5 cos(t)
correct approach (accept any variable for t) (A1)
eg −13 sin(t + 0.395), sketch of d′, (1.18, −13), t = 4.32
maximum speed = 13 (cms−1) A1 N2
[3 marks]
recognizing that acceleration is needed (M1)
eg a(t), d "(t)
correct equation (accept any variable for t) (A1)
eg
valid attempt to solve their equation (M1)
eg sketch, 1.33
1.02154
1.02 A2 N3
[5 marks]
Examiners report
Question
In a company it is found that 25 % of the employees encountered traffic on their way to work. From those who encountered traffic the probability of being late for work is 80 %.
From those who did not encounter traffic, the probability of being late for work is 15 %.
The tree diagram illustrates the information.
The company investigates the different means of transport used by their employees in the past year to travel to work. It was found that the three most common means of transport used to travel to work were public transportation (P ), car (C ) and bicycle (B ).
The company finds that 20 employees travelled by car, 28 travelled by bicycle and 19 travelled by public transportation in the last year.
Some of the information is shown in the Venn diagram.
There are 54 employees in the company.
Write down the value of a.
Write down the value of b.
Use the tree diagram to find the probability that an employee encountered traffic and was late for work.
Use the tree diagram to find the probability that an employee was late for work.
Use the tree diagram to find the probability that an employee encountered traffic given that they were late for work.
Find the value of x.
Find the value of y.
Find the number of employees who, in the last year, did not travel to work by car, bicycle or public transportation.
Find .
Markscheme
a = 0.2 (A1)
[1 mark]
b = 0.85 (A1)
[1 mark]
0.25 × 0.8 (M1)
Note: Award (M1) for a correct product.
(A1)(G2)
[2 marks]
0.25 × 0.8 + 0.75 × 0.15 (A1)(ft)(M1)
Note: Award (A1)(ft) for their (0.25 × 0.8) and (0.75 × 0.15), (M1) for adding two products.
(A1)(ft)(G3)
Note: Award the final (A1)(ft) only if answer does not exceed 1. Follow through from part (b)(i).
[3 marks]
(A1)(ft)(A1)(ft)
Note: Award (A1)(ft) for a correct numerator (their part (b)(i)), (A1)(ft) for a correct denominator (their part (b)(ii)). Follow through from parts (b)(i) and (b)(ii).
(A1)(ft)(G3)
Note: Award final (A1)(ft) only if answer does not exceed 1.
[3 marks]
(x =) 3 (A1)
[1 Mark]
(y =) 10 (A1)(ft)
Note: Following through from part (c)(i) but only if their x is less than or equal to 13.
[1 Mark]
54 − (10 + 3 + 4 + 2 + 6 + 8 + 13) (M1)
Note: Award (M1) for subtracting their correct sum from 54. Follow through from their part (c).
= 8 (A1)(ft)(G2)
Note: Award (A1)(ft) only if their sum does not exceed 54. Follow through from their part (c).
[2 marks]
6 + 8 + 13 (M1)
Note: Award (M1) for summing 6, 8 and 13.
27 (A1)(G2)
[2 marks]
Examiners report
Question
On one day 180 flights arrived at a particular airport. The distance travelled and the arrival status for each incoming flight was recorded. The flight was then classified as on time, slightly delayed, or heavily delayed.
The results are shown in the following table.
A χ2 test is carried out at the 10 % significance level to determine whether the arrival status of incoming flights is independent of the distance travelled.
The critical value for this test is 7.779.
A flight is chosen at random from the 180 recorded flights.
State the alternative hypothesis.
Calculate the expected frequency of flights travelling at most 500 km and arriving slightly delayed.
Write down the number of degrees of freedom.
Write down the χ2 statistic.
Write down the associated p-value.
State, with a reason, whether you would reject the null hypothesis.
Write down the probability that this flight arrived on time.
Given that this flight was not heavily delayed, find the probability that it travelled between 500 km and 5000 km.
Two flights are chosen at random from those which were slightly delayed.
Find the probability that each of these flights travelled at least 5000 km.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
The arrival status is dependent on the distance travelled by the incoming flight (A1)
Note: Accept “associated” or “not independent”.
[1 mark]
OR (M1)
Note: Award (M1) for correct substitution into expected value formula.
= 15 (A1) (G2)
[2 marks]
4 (A1)
Note: Award (A0) if “2 + 2 = 4” is seen.
[1 mark]
9.55 (9.54671…) (G2)
Note: Award (G1) for an answer of 9.54.
[2 marks]
0.0488 (0.0487961…) (G1)
[1 mark]
Reject the Null Hypothesis (A1)(ft)
Note: Follow through from their hypothesis in part (a).
9.55 (9.54671…) > 7.779 (R1)(ft)
OR
0.0488 (0.0487961…) < 0.1 (R1)(ft)
Note: Do not award (A1)(ft)(R0)(ft). Follow through from part (d). Award (R1)(ft) for a correct comparison, (A1)(ft) for a consistent conclusion with the answers to parts (a) and (d). Award (R1)(ft) for χ2calc > χ2crit , provided the calculated value is explicitly seen in part (d)(i).
[2 marks]
(A1)(A1) (G2)
Note: Award (A1) for correct numerator, (A1) for correct denominator.
[2 marks]
(A1)(A1) (G2)
Note: Award (A1) for correct numerator, (A1) for correct denominator.
[2 marks]
(A1)(M1)
Note: Award (A1) for two correct fractions and (M1) for multiplying their two fractions.
(A1) (G2)
[3 marks]
Examiners report
Question
Consider the function , where x > 0 and k is a constant.
The graph of the function passes through the point with coordinates (4 , 2).
P is the minimum point of the graph of f (x).
Sketch the graph of y = f (x) for 0 < x ≤ 6 and −30 ≤ y ≤ 60.
Clearly indicate the minimum point P and the x-intercepts on your graph.
Markscheme
(A1)(A1)(ft)(A1)(ft)(A1)(ft)
Note: Award (A1) for correct window. Axes must be labelled.
(A1)(ft) for a smooth curve with correct shape and zeros in approximately correct positions relative to each other.
(A1)(ft) for point P indicated in approximately the correct position. Follow through from their x-coordinate in part (c). (A1)(ft) for two x-intercepts identified on the graph and curve reflecting asymptotic properties.
[4 marks]
Examiners report
Question
Contestants in a TV gameshow try to get through three walls by passing through doors without falling into a trap. Contestants choose doors at random.
If they avoid a trap they progress to the next wall.
If a contestant falls into a trap they exit the game before the next contestant plays.
Contestants are not allowed to watch each other attempt the game.
The first wall has four doors with a trap behind one door.
Ayako is a contestant.
Natsuko is the second contestant.
The second wall has five doors with a trap behind two of the doors.
The third wall has six doors with a trap behind three of the doors.
The following diagram shows the branches of a probability tree diagram for a contestant in the game.
Write down the probability that Ayako avoids the trap in this wall.
Find the probability that only one of Ayako and Natsuko falls into a trap while attempting to pass through a door in the first wall.
Copy the probability tree diagram and write down the relevant probabilities along the branches.
A contestant is chosen at random. Find the probability that this contestant fell into a trap while attempting to pass through a door in the second wall.
A contestant is chosen at random. Find the probability that this contestant fell into a trap.
120 contestants attempted this game.
Find the expected number of contestants who fell into a trap while attempting to pass through a door in the third wall.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(0.75, 75%) (A1)
[1 mark]
OR (M1)(M1)
Note: Award (M1) for their product seen, and (M1) for adding their two products or multiplying their product by 2.
(A1)(ft) (G3)
Note: Follow through from part (a), but only if the sum of their two fractions is 1.
[3 marks]
(A1)(ft)(A1)(A1)
Note: Award (A1) for each correct pair of branches. Follow through from part (a).
[3 marks]
(M1)
Note: Award (M1) for correct probabilities multiplied together.
(A1)(ft) (G2)
Note: Follow through from their tree diagram or part (a).
[2 marks]
OR (M1)(M1)
Note: Award (M1) for and (M1) for subtracting their correct probability from 1, or adding to their .
(A1)(ft) (G2)
Note: Follow through from their tree diagram.
[3 marks]
(M1)(M1)
Note: Award (M1) for and (M1) for multiplying by 120.
= 27 (A1)(ft) (G3)
Note: Follow through from their tree diagram or their from their calculation in part (d)(ii).
[3 marks]
Examiners report
Question
Consider the curve y = 2x3 − 9x2 + 12x + 2, for −1 < x < 3
Sketch the curve for −1 < x < 3 and −2 < y < 12.
A teacher asks her students to make some observations about the curve.
Three students responded.
Nadia said “The x-intercept of the curve is between −1 and zero”.
Rick said “The curve is decreasing when x < 1 ”.
Paula said “The gradient of the curve is less than zero between x = 1 and x = 2 ”.
State the name of the student who made an incorrect observation.
Find .
Show that the stationary points of the curve are at x = 1 and x = 2.
Given that y = 2x3 − 9x2 + 12x + 2 = k has three solutions, find the possible values of k.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)(A1)(A1)(A1)
Note: Award (A1) for correct window (condone a window which is slightly off) and axes labels. An indication of window is necessary. −1 to 3 on the x-axis and −2 to 12 on the y-axis and a graph in that window.
(A1) for correct shape (curve having cubic shape and must be smooth).
(A1) for both stationary points in the 1st quadrant with approximate correct position,
(A1) for intercepts (negative x-intercept and positive y intercept) with approximate correct position.
[4 marks]
Rick (A1)
Note: Award (A0) if extra names stated.
[1 mark]
6x2 − 18x + 12 (A1)(A1)(A1)
Note: Award (A1) for each correct term. Award at most (A1)(A1)(A0) if extra terms seen.
[3 marks]
6x2 − 18x + 12 = 0 (M1)
Note: Award (M1) for equating their derivative to 0. If the derivative is not explicitly equated to 0, but a subsequent solving of their correct equation is seen, award (M1).
6(x − 1)(x − 2) = 0 (or equivalent) (M1)
Note: Award (M1) for correct factorization. The final (M1) is awarded only if answers are clearly stated.
Award (M0)(M0) for substitution of 1 and of 2 in their derivative.
x = 1, x = 2 (AG)
[2 marks]
6 < k < 7 (A1)(A1)(ft)(A1)
Note: Award (A1) for an inequality with 6, award (A1)(ft) for an inequality with 7 from their part (c) provided it is greater than 6, (A1) for their correct strict inequalities. Accept ]6, 7[ or (6, 7).
[3 marks]
Examiners report
Question
A manufacturer makes trash cans in the form of a cylinder with a hemispherical top. The trash can has a height of 70 cm. The base radius of both the cylinder and the hemispherical top is 20 cm.
A designer is asked to produce a new trash can.
The new trash can will also be in the form of a cylinder with a hemispherical top.
This trash can will have a height of H cm and a base radius of r cm.
There is a design constraint such that H + 2r = 110 cm.
The designer has to maximize the volume of the trash can.
Write down the height of the cylinder.
Find the total volume of the trash can.
Find the height of the cylinder, h , of the new trash can, in terms of r.
Show that the volume, V cm3 , of the new trash can is given by
.
Using your graphic display calculator, find the value of r which maximizes the value of V.
The designer claims that the new trash can has a capacity that is at least 40% greater than the capacity of the original trash can.
State whether the designer’s claim is correct. Justify your answer.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
50 (cm) (A1)
[1 mark]
(M1)(M1)(M1)
Note: Award (M1) for their correctly substituted volume of cylinder, (M1) for correctly substituted volume of sphere formula, (M1) for halving the substituted volume of sphere formula. Award at most (M1)(M1)(M0) if there is no addition of the volumes.
(A1)(ft) (G3)
Note: Follow through from part (a).
[4 marks]
h = H − r (or equivalent) OR H = 110 − 2r (M1)
Note: Award (M1) for writing h in terms of H and r or for writing H in terms of r.
(h =) 110 − 3r (A1) (G2)
[2 marks]
(M1)(M1)(M1)
Note: Award (M1) for volume of hemisphere, (M1) for correct substitution of their h into the volume of a cylinder, (M1) for addition of two correctly substituted volumes leading to the given answer. Award at most (M1)(M1)(M0) for subsequent working that does not lead to the given answer. Award at most (M1)(M1)(M0) for substituting H = 110 − 2r as their h.
(AG)
[3 marks]
(r =) 31.4 (cm) (31.4285… (cm)) (G2)
OR
(M1)
Note: Award (M1) for setting the correct derivative equal to zero.
(r =) 31.4 (cm) (31.4285… (cm)) (A1)
[2 marks]
(M1)
Note: Award (M1) for correct substitution of their 31.4285… into the given equation.
= 114000 (113781…) (A1)(ft)
Note: Follow through from part (e).
(increase in capacity =) (R1)(ft)
Note: Award (R1)(ft) for finding the correct percentage increase from their two volumes.
OR
1.4 × 79587.0… = 111421.81… (R1)(ft)
Note: Award (R1)(ft) for finding the capacity of a trash can 40% larger than the original.
Claim is correct (A1)(ft)
Note: Follow through from parts (b), (e) and within part (f). The final (R1)(A1)(ft) can be awarded for their correct reason and conclusion. Do not award (R0)(A1)(ft).
[4 marks]
Examiners report
Question
A particle moves along a straight line so that its velocity, m s−1, after seconds is given by , for 0 ≤ ≤ 5.
Find when the particle is at rest.
Find the acceleration of the particle when .
Find the total distance travelled by the particle.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach (M1)
eg , sketch of graph
2.95195
(exact), (s) A1 N2
[2 marks]
valid approach (M1)
eg ,
0.659485
= 1.96 ln 1.4 (exact), = 0.659 (m s−2) A1 N2
[2 marks]
correct approach (A1)
eg ,
5.3479
distance = 5.35 (m) A2 N3
[3 marks]
Examiners report
Question
All lengths in this question are in metres.
Consider the function , for −2 ≤ ≤ 2. In the following diagram, the shaded region is enclosed by the graph of and the -axis.
A container can be modelled by rotating this region by 360˚ about the -axis.
Water can flow in and out of the container.
The volume of water in the container is given by the function , for 0 ≤ ≤ 4 , where is measured in hours and is measured in m3. The rate of change of the volume of water in the container is given by .
The volume of water in the container is increasing only when < < .
Find the volume of the container.
Find the value of and of .
During the interval < < , he volume of water in the container increases by m3. Find the value of .
When = 0, the volume of water in the container is 2.3 m3. It is known that the container is never completely full of water during the 4 hour period.
Find the minimum volume of empty space in the container during the 4 hour period.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to substitute correct limits or the function into formula involving (M1)
eg ,
4.18879
volume = 4.19, (exact) (m3) A2 N3
Note: If candidates have their GDC incorrectly set in degrees, award M marks where appropriate, but no A marks may be awarded. Answers from degrees are p = 13.1243 and q = 26.9768 in (b)(i) and 12.3130 or 28.3505 in (b)(ii).
[3 marks]
recognizing the volume increases when is positive (M1)
eg > 0, sketch of graph of indicating correct interval
1.73387, 3.56393
= 1.73, = 3.56 A1A1 N3
[3 marks]
valid approach to find change in volume (M1)
eg ,
3.74541
total amount = 3.75 (m3) A2 N3
[3 marks]
Note: There may be slight differences in the final answer, depending on which values candidates carry through from previous parts. Accept answers that are consistent with correct working.
recognizing when the volume of water is a maximum (M1)
eg maximum when ,
valid approach to find maximum volume of water (M1)
eg , , 3.85745
correct expression for the difference between volume of container and maximum value (A1)
eg , 4.19 − 3.85745
0.331334
0.331 (m3) A2 N3
[5 marks]
Examiners report
Question
Consider the function .
Sketch the graph of y = f (x), for −4 ≤ x ≤ 3 and −50 ≤ y ≤ 100.
Use your graphic display calculator to find the equation of the tangent to the graph of y = f (x) at the point (–2, 38.75).
Give your answer in the form y = mx + c.
Sketch the graph of the function g (x) = 10x + 40 on the same axes.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)(A1)(A1)(A1)
Note: Award (A1) for axis labels and some indication of scale; accept y or f(x).
Use of graph paper is not required. If no scale is given, assume the given window for zero and minimum point.
Award (A1) for smooth curve with correct general shape.
Award (A1) for x-intercept closer to y-axis than to end of sketch.
Award (A1) for correct local minimum with x-coordinate closer to y-axis than end of sketch and y-coordinate less than half way to top of sketch.
Award at most (A1)(A0)(A1)(A1) if the sketch intersects the y-axis or if the sketch curves away from the y-axis as x approaches zero.
[4 marks]
y = −9.25x + 20.3 (y = −9.25x + 20.25) (A1)(A1)
Note: Award (A1) for −9.25x, award (A1) for +20.25, award a maximum of (A0)(A1) if answer is not an equation.
[2 marks]
correct line, y = 10x + 40, seen on sketch (A1)(A1)
Note: Award (A1) for straight line with positive gradient, award (A1) for x-intercept and y-intercept in approximately the correct positions. Award at most (A0)(A1) if ruler not used. If the straight line is drawn on different axes to part (a), award at most (A0)(A1).
[2 marks]
Examiners report
Question
Haruka has an eco-friendly bag in the shape of a cuboid with width 12 cm, length 36 cm and height of 9 cm. The bag is made from five rectangular pieces of cloth and is open at the top.
Nanako decides to make her own eco-friendly bag in the shape of a cuboid such that the surface area is minimized.
The width of Nanako’s bag is x cm, its length is three times its width and its height is y cm.
The volume of Nanako’s bag is 3888 cm3.
Calculate the area of cloth, in cm2, needed to make Haruka’s bag.
Calculate the volume, in cm3, of the bag.
Use this value to write down, and simplify, the equation in x and y for the volume of Nanako’s bag.
Write down and simplify an expression in x and y for the area of cloth, A, used to make Nanako’s bag.
Use your answers to parts (c) and (d) to show that
.
Find .
Use your answer to part (f) to show that the width of Nanako’s bag is 12 cm.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
36 × 12 + 2(9 ×12) + 2(9 × 36) (M1)
Note: Award (M1) for correct substitution into surface area of cuboid formula.
= 1300 (cm2) (1296 (cm2)) (A1)(G2)
[2 marks]
36 × 9 ×12 (M1)
Note: Award (M1) for correct substitution into volume of cuboid formula.
= 3890 (cm3) (3888 (cm3)) (A1)(G2)
[2 marks]
3 × × = 3888 (M1)
Note: Award (M1) for correct substitution into volume of cuboid formula and equated to 3888.
2 = 1296 (A1)(G2)
Note: Award (A1) for correct fully simplified volume of cuboid.
Accept .
[2 marks]
(A =) 3x2 + 2(xy) + 2(3xy) (M1)
Note: Award (M1) for correct substitution into surface area of cuboid formula.
(A =) 3x2 + 8xy (A1)(G2)
Note: Award (A1) for correct simplified surface area of cuboid formula.
[2 marks]
(A1)(ft)(M1)
Note: Award (A1)(ft) for correct rearrangement of their part (c) seen (rearrangement may be seen in part(c)), award (M1) for substitution of their part (c) into their part (d) but only if this leads to the given answer, which must be shown.
(AG)
[2 marks]
(A1)(A1)(A1)
Note: Award (A1) for , (A1) for −10368, (A1) for . Award a maximum of (A1)(A1)(A0) if any extra terms seen.
[3 marks]
(M1)
Note: Award (M1) for equating their to zero.
OR OR (M1)
Note: Award (M1) for correctly rearranging their equation so that fractions are removed.
(A1)
(cm) (AG)
Note: The (AG) line must be seen for the final (A1) to be awarded. Substituting invalidates the method, award a maximum of (M1)(M0)(A0).
[3 marks]
Examiners report
Question
160 students attend a dual language school in which the students are taught only in Spanish or taught only in English.
A survey was conducted in order to analyse the number of students studying Biology or Mathematics. The results are shown in the Venn diagram.
Set S represents those students who are taught in Spanish.
Set B represents those students who study Biology.
Set M represents those students who study Mathematics.
A student from the school is chosen at random.
Find the number of students in the school that are taught in Spanish.
Find the number of students in the school that study Mathematics in English.
Find the number of students in the school that study both Biology and Mathematics.
Write down .
Write down .
Find the probability that this student studies Mathematics.
Find the probability that this student studies neither Biology nor Mathematics.
Find the probability that this student is taught in Spanish, given that the student studies Biology.
Markscheme
10 + 40 + 28 + 17 (M1)
= 95 (A1)(G2)
Note: Award (M1) for each correct sum (for example: 10 + 40 + 28 + 17) seen.
[2 marks]
20 + 12 (M1)
= 32 (A1)(G2)
Note: Award (M1) for each correct sum (for example: 10 + 40 + 28 + 17) seen.
[2 marks]
12 + 40 (M1)
= 52 (A1)(G2)
Note: Award (M1) for each correct sum (for example: 10 + 40 + 28 + 17) seen.
[2 marks]
78 (A1)
[1 mark]
12 (A1)
[1 mark]
(A1)(A1) (G2)
Note: Throughout part (c), award (A1) for correct numerator, (A1) for correct denominator. All answers must be probabilities to award (A1).
[2 marks]
(A1)(A1) (G2)
Note: Throughout part (c), award (A1) for correct numerator, (A1) for correct denominator. All answers must be probabilities to award (A1).
[2 marks]
(A1)(A1) (G2)
Note: Throughout part (c), award (A1) for correct numerator, (A1) for correct denominator. All answers must be probabilities to award (A1).
[2 marks]
Examiners report
Question
Let . The following diagram shows part of the graph of .
Find the -intercept of the graph of .
The region enclosed by the graph of , the -axis and the -axis is rotated 360º about the -axis. Find the volume of the solid formed.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach (M1)
eg ,
0.693147
= ln 2 (exact), 0.693 A1 N2
[2 marks]
attempt to substitute either their correct limits or the function into formula (M1)
involving
eg , ,
3.42545
volume = 3.43 A2 N3
[3 marks]
Examiners report
Question
Consider the function , .
The graph of has a horizontal tangent line at and at . Find .
Markscheme
valid method (M1)
eg , ,
(accept ) A1 N2
[2 marks]
Examiners report
Question
Let , for 0 ≤ ≤ 1.
Sketch the graph of on the grid below:
Find the -coordinates of the points of inflexion of the graph of .
Hence find the values of for which the graph of is concave-down.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
A1A1A1 N3
Note: Only if the shape is approximately correct with exactly 2 maximums and 1 minimum on the interval 0 ≤ ≤ 0, award the following:
A1 for correct domain with both endpoints within circle and oval.
A1 for passing through the other -intercepts within the circles.
A1 for passing through the three turning points within circles (ignore -intercepts and extrema outside of the domain).
[3 marks]
evidence of reasoning (may be seen on graph) (M1)
eg , (0.524, 0), (0.785, 0)
0.523598, 0.785398
, A1A1 N3
Note: Award M1A1A0 if any solution outside domain (eg ) is also included.
[3 marks]
A2 N2
Note: Award A1 if any correct interval outside domain also included, unless additional solutions already penalized in (b).
Award A0 if any incorrect intervals are also included.
[2 marks]
Examiners report
Question
The population of fish in a lake is modelled by the function
, 0 ≤ ≤ 30 , where is measured in months.
Find the population of fish at = 10.
Find the rate at which the population of fish is increasing at = 10.
Find the value of for which the population of fish is increasing most rapidly.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach (M1)
eg (10)
235.402
235 (fish) (must be an integer) A1 N2
[2 marks]
recognizing rate of change is derivative (M1)
eg rate = , (10) , sketch of , 35 (fish per month)
35.9976
36.0 (fish per month) A1 N2
[2 marks]
valid approach (M1)
eg maximum of , = 0
15.890
15.9 (months) A1 N2
[2 marks]
Examiners report
Question
In this question distance is in centimetres and time is in seconds.
Particle A is moving along a straight line such that its displacement from a point P, after seconds, is given by , 0 ≤ ≤ 25. This is shown in the following diagram.
Another particle, B, moves along the same line, starting at the same time as particle A. The velocity of particle B is given by , 0 ≤ ≤ 25.
Find the initial displacement of particle A from point P.
Find the value of when particle A first reaches point P.
Find the value of when particle A first changes direction.
Find the total distance travelled by particle A in the first 3 seconds.
Given that particles A and B start at the same point, find the displacement function for particle B.
Find the other value of when particles A and B meet.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach (M1)
eg
15 (cm) A1 N2
[2 marks]
valid approach (M1)
eg
2.46941
= 2.47 (seconds) A1 N2
[2 marks]
recognizing when change in direction occurs (M1)
eg slope of changes sign, , minimum point, 10.0144, (4.08, −4.66)
4.07702
= 4.08 (seconds) A1 N2
[2 marks]
METHOD 1 (using displacement)
correct displacement or distance from P at (seen anywhere) (A1)
eg −2.69630, 2.69630
valid approach (M1)
eg 15 + 2.69630, , −17.6963
17.6963
17.7 (cm) A1 N2
METHOD 2 (using velocity)
attempt to substitute either limits or the velocity function into distance formula involving (M1)
eg ,
17.6963
17.7 (cm) A1 N2
[3 marks]
recognize the need to integrate velocity (M1)
eg
(accept instead of and missing ) (A2)
substituting initial condition into their integrated expression (must have ) (M1)
eg ,
A1 N3
[5 marks]
valid approach (M1)
eg , sketch, (9.30404, 2.86710)
9.30404
(seconds) A1 N2
Note: If candidates obtain in part (e)(i), there are 2 solutions for part (e)(ii), 1.32463 and 7.79009. Award the last A1 in part (e)(ii) only if both solutions are given.
[2 marks]
Examiners report
Question
Let . The line is tangent to the graph of at .
can be expressed in the form r u.
The direction vector of is .
Find the gradient of .
Find u.
Find the acute angle between and .
Find .
Hence, write down .
Hence or otherwise, find the obtuse angle formed by the tangent line to at and the tangent line to at .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to find (M1)
eg , ,
−0.25 (exact) A1 N2
[2 marks]
u null or any scalar multiple A2 N2
[2 marks]
correct scalar product and magnitudes (A1)(A1)(A1)
scalar product
magnitudes ,
substitution of their values into correct formula (M1)
eg , , 2.1112, 120.96°
1.03037 , 59.0362°
angle = 1.03 , 59.0° A1 N4
[5 marks]
attempt to form composite (M1)
eg , ,
correct working (A1)
eg ,
A1 N2
[3 marks]
(accept , ) A1 N1
Note: Award A0 in part (ii) if part (i) is incorrect.
Award A0 in part (ii) if the candidate has found by interchanging and .
[1 mark]
METHOD 1
recognition of symmetry about (M1)
eg (2, 8) ⇔ (8, 2)
evidence of doubling their angle (M1)
eg ,
2.06075, 118.072°
2.06 (radians) (118 degrees) A1 N2
METHOD 2
finding direction vector for tangent line at (A1)
eg ,
substitution of their values into correct formula (must be from vectors) (M1)
eg ,
2.06075, 118.072°
2.06 (radians) (118 degrees) A1 N2
METHOD 3
using trigonometry to find an angle with the horizontal (M1)
eg ,
finding both angles of rotation (A1)
eg ,
2.06075, 118.072°
2.06 (radians) (118 degrees) A1 N2
[3 marks]
Examiners report
Question
Sila High School has 110 students. They each take exactly one language class from a choice of English, Spanish or Chinese. The following table shows the number of female and male students in the three different language classes.
A test was carried out at the 5 % significance level to analyse the relationship between gender and student choice of language class.
Use your graphic display calculator to write down
The critical value at the 5 % significance level for this test is 5.99.
One student is chosen at random from this school.
Another student is chosen at random from this school.
Write down the null hypothesis, H0 , for this test.
State the number of degrees of freedom.
the expected frequency of female students who chose to take the Chinese class.
the statistic.
State whether or not H0 should be rejected. Justify your statement.
Find the probability that the student does not take the Spanish class.
Find the probability that neither of the two students take the Spanish class.
Find the probability that at least one of the two students is female.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(H0:) (choice of) language is independent of gender (A1)
Note: Accept “there is no association between language (choice) and gender”. Accept “language (choice) is not dependent on gender”. Do not accept “not related” or “not correlated” or “not influenced”.
[1 mark]
2 (AG)
[1 mark]
16.4 (16.4181…) (G1)
[1 mark]
(8.68507…) (G2)
[2 marks]
(we) reject the null hypothesis (A1)(ft)
8.68507… > 5.99 (R1)(ft)
Note: Follow through from part (c)(ii). Accept “do not accept” in place of “reject.” Do not award (A1)(ft)(R0).
OR
(we) reject the null hypothesis (A1)
0.0130034 < 0.05 (R1)
Note: Accept “do not accept” in place of “reject.” Do not award (A1)(ft)(R0).
[2 marks]
(A1)(A1)(G2)
Note: Award (A1) for correct numerator, (A1) for correct denominator.
[2 marks]
(M1)(M1)
Note: Award (M1) for multiplying two fractions. Award (M1) for multiplying their correct fractions.
OR
(M1)(M1)
Note: Award (M1) for correct products; (M1) for adding 4 products.
(A1)(ft)(G2)
Note: Follow through from their answer to part (e)(i).
[3 marks]
(M1)(M1)
Note: Award (M1) for multiplying two correct fractions. Award (M1) for subtracting their product of two fractions from 1.
OR
(M1)(M1)
Note: Award (M1) for correct products; (M1) for adding three products.
(A1)(G2)
[3 marks]
Examiners report
Question
Consider the function .
The function has one local maximum at and one local minimum at .
Write down the -intercept of the graph of .
Sketch the graph of for −3 ≤ ≤ 3 and −4 ≤ ≤ 12.
Determine the range of for ≤ ≤ .
Markscheme
−1 (A1)
Note: Accept (0, −1).
[1 mark]
(A1)(A1)(A1)(A1)
Note: Award (A1) for correct window and axes labels, −3 to 3 should be indicated on the -axis and −4 to 12 on the -axis.
(A1)) for smooth curve with correct cubic shape;
(A1) for -intercepts: one close to −3, the second between −1 and 0, and third between 1 and 2; and -intercept at approximately −1;
(A1) for local minimum in the 4th quadrant and maximum in the 2nd quadrant, in approximately correct positions.
Graph paper does not need to be used. If window not given award at most (A0)(A1)(A0)(A1).
[4 marks]
(A1)(ft)(A1)(ft)(A1)
Note: Award (A1) for −1.27 seen, (A1) for 1.33 seen, and (A1) for correct weak inequalities with their endpoints in the correct order. For example, award (A0)(A0)(A0) for answers like . Accept in place of . Accept alternative correct notation such as [−1.27, 1.33].
Follow through from their and values from part (g) only if their and values are between −4 and 12. Award (A0)(A0)(A0) if their values from (g) are given as the endpoints.
[3 marks]
Examiners report
Question
Let , and .
Find .
Let be a point on the graph of . The tangent to the graph of at is parallel to the graph of .
Find the -coordinate of .
Markscheme
attempt to form composite (in any order) (M1)
eg ,
A1 N2
[2 marks]
recognizing that the gradient of the tangent is the derivative (M1)
eg
correct derivative (seen anywhere) (A1)
correct value for gradient of (seen anywhere) (A1)
,
setting their derivative equal to (M1)
(exact), A1 N3
[5 marks]
Examiners report
Question
Let , for . The following diagram shows the graph of .
There are -intercepts at and at . There is a maximum at point where , and a point of inflexion at point where .
Find the value of .
Write down the coordinates of .
Find the equation of the tangent to the graph of at .
Find the coordinates of .
Find the rate of change of at .
Let be the region enclosed by the graph of , the -axis and the lines and . The region is rotated 360º about the -axis. Find the volume of the solid formed.
Markscheme
evidence of valid approach (M1)
eg ,
A1 N2
[2 marks]
,
A2 N2
[2 marks]
valid approach (M1)
eg tangent at maximum point is horizontal,
(must be an equation) A1 N2
[2 marks]
METHOD 1 (using GDC)
valid approach M1
eg , max/min on ,
sketch of either or , with max/min or root (respectively) (A1)
A1 N1
substituting their value into (M1)
eg
(exact) (accept ) A1 N1
METHOD 2 (analytical)
A1
valid approach (M1)
eg ,
A1 N1
substituting their value into (M1)
eg
(exact) (accept ) A1 N1
[5 marks]
recognizing rate of change is (M1)
eg ,
rate of change is (exact) A1 N2
[2 marks]
attempt to substitute either their limits or the function into volume formula (M1)
eg , ,
volume A2 N3
[3 marks]
Examiners report
Question
A rocket is travelling in a straight line, with an initial velocity of m s−1. It accelerates to a new velocity of m s−1 in two stages.
During the first stage its acceleration, m s−2, after seconds is given by , where .
The first stage continues for seconds until the velocity of the rocket reaches m s−1.
Find an expression for the velocity, m s−1, of the rocket during the first stage.
Find the distance that the rocket travels during the first stage.
During the second stage, the rocket accelerates at a constant rate. The distance which the rocket travels during the second stage is the same as the distance it travels during the first stage.
Find the total time taken for the two stages.
Markscheme
recognizing that (M1)
correct integration A1
eg
attempt to find using their (M1)
eg
A1 N3
[4 marks]
evidence of valid approach to find time taken in first stage (M1)
eg graph,
A1
attempt to substitute their and/or their limits into distance formula (M1)
eg , ,
distance is (m) A1 N3
[4 marks]
recognizing velocity of second stage is linear (seen anywhere) R1
eg graph, ,
valid approach (M1)
eg
correct equation (A1)
eg
time for stage two ( from 3 sf) A2
( from 3 sf)
seconds ( from 3 sf) A1 N3
[6 marks]
Examiners report
Question
Consider a function , for . The derivative of is given by .
The graph of is concave-down when .
Show that .
Find the least value of .
Find .
Let be the region enclosed by the graph of , the -axis and the lines and . The area of is , correct to three significant figures.
Find .
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
evidence of choosing the quotient rule (M1)
eg
derivative of is (must be seen in rule) (A1)
derivative of is (must be seen in rule) (A1)
correct substitution into the quotient rule A1
eg
AG N0
METHOD 2
evidence of choosing the product rule (M1)
eg
derivative of is (must be seen in rule) (A1)
derivative of is (must be seen in rule) (A1)
correct substitution into the product rule A1
eg
AG N0
[4 marks]
METHOD 1 (2nd derivative) (M1)
valid approach
eg
(exact) A1 N2
METHOD 2 (1st derivative)
valid attempt to find local maximum on (M1)
eg sketch with max indicated,
(exact) A1 N2
[2 marks]
evidence of valid approach using substitution or inspection (M1)
eg
A2 N3
[3 marks]
recognizing that area (seen anywhere) (M1)
recognizing that their answer to (c) is their (accept absence of ) (M1)
eg
correct value for (seen anywhere) (A1)
eg
correct integration for (seen anywhere) (A1)
adding their integrated expressions and equating to (do not accept an expression which involves an integral) (M1)
eg
(A1)
A1 N4
[7 marks]
Examiners report
Question
Hyungmin designs a concrete bird bath. The bird bath is supported by a pedestal. This is shown in the diagram.
The interior of the bird bath is in the shape of a cone with radius , height and a constant slant height of .
Let be the volume of the bird bath.
Hyungmin wants the bird bath to have maximum volume.
Write down an equation in and that shows this information.
Show that .
Find .
Using your answer to part (c), find the value of for which is a maximum.
Find the maximum volume of the bird bath.
To prevent leaks, a sealant is applied to the interior surface of the bird bath.
Find the surface area to be covered by the sealant, given that the bird bath has maximum volume.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(or equivalent) (A1)
Note: Accept equivalent expressions such as or . Award (A0) for a final answer of or , or any further incorrect working.
[1 mark]
OR (M1)
Note: Award (M1) for correct substitution in the volume of cone formula.
(AG)
Note: The final line must be seen, with no incorrect working, for the (M1) to be awarded.
[1 mark]
(A1)(A1)
Note: Award (A1) for , (A1) for . Award at most (A1)(A0) if extra terms are seen. Award (A0) for the term .
[2 marks]
(M1)
Note: Award (M1) for equating their derivative to zero. Follow through from part (c).
OR
sketch of (M1)
Note: Award (M1) for a labelled sketch of with the curve/axes correctly labelled or the -intercept explicitly indicated.
(A1)(ft)
Note: An unsupported is awarded no marks. Graphing the function is not an acceptable method and (M0)(A0) should be awarded. Follow through from part (c). Given the restraints of the question, is not possible.
[2 marks]
(M1)
OR
(M1)
Note: Award (M1) for substituting their in the volume formula.
(A1)(ft)(G2)
Note: Follow through from part (d).
[2 marks]
(A1)(ft)(M1)
Note: Award (A1) for their correct radius seen .
Award (M1) for correctly substituted curved surface area formula for a cone.
(A1)(ft)(G2)
Note: Follow through from parts (a) and (d).
[3 marks]
Examiners report
Question
Find .
Given and , find .
Markscheme
correct integration A1A1A1
Note: Award A1 for , A1 for and A1 for
[3 marks]
recognition that (M1)
(A1)
A1
[3 marks]
Examiners report
Question
A particle moves in a straight line. The velocity, , of the particle at time seconds is given by , for .
The following diagram shows the graph of .
Find the smallest value of for which the particle is at rest.
Find the total distance travelled by the particle.
Find the acceleration of the particle when .
Markscheme
recognising (M1)
(sec) A1
Note: Do not award A1 if additional values are given.
[2 marks]
OR (A1)
A1
[2 marks]
recognizing acceleration at is given by (M1)
acceleration
A1
[2 marks]
Examiners report
Question
Consider the function defined by for .
The graph of and the line intersect at point .
The line has a gradient of and is a tangent to the graph of at the point .
The shaded region is enclosed by the graph of and the lines and .
Find the -coordinate of .
Find the exact coordinates of .
Show that the equation of is .
Find the -coordinate of the point where intersects the line .
Hence, find the area of .
The line is tangent to the graphs of both and the inverse function .
Find the shaded area enclosed by the graphs of and and the line .
Markscheme
Attempt to find the point of intersection of the graph of and the line (M1)
A1
[2 marks]
A1
attempt to set the gradient of equal to (M1)
has coordinates (accept () A1A1
Note: Award A1 for each value, even if the answer is not given as a coordinate pair.
Do not accept or as a final value for . Do not accept or as a final value for .
[4 marks]
attempt to substitute coordinates of (in any order) into an appropriate equation (M1)
OR A1
equation of is AG
[2 marks]
A1
[1 mark]
appropriate method to find the sum of two areas using integrals of the difference of two functions (M1)
Note: Allow absence of incorrect limits.
(A1)(A1)
Note: Award A1 for one correct integral expression including correct limits and integrand.
Award A1 for a second correct integral expression including correct limits and integrand.
A1
[4 marks]
by symmetry (M1)
A1
Note: Accept any answer that rounds to (but do not accept ).
[2 marks]
Examiners report
Question
The derivative of a function is given by , where . The graph of passes through the point . Find .
Markscheme
METHOD 1
recognises that (M1)
(A1)(A1)
Note: Award A1 for each integrated term.
substitutes and into their integrated function (must involve ) (M1)
A1
METHOD 2
attempts to write both sides in the form of a definite integral (M1)
(A1)
(A1)(A1)
Note: Award A1 for and A1 for .
A1
[5 marks]
Examiners report
While many students were successful in solving this question, some did not consider the constant of integration or struggled to integrate the exponential term. A few students lost the final mark for stopping at and not giving the formula for .
Question
A particle moves along a straight line so that its velocity, , after seconds is given by for .
Find the value of when the particle is at rest.
Find the acceleration of the particle when it changes direction.
Find the total distance travelled by the particle.
Markscheme
recognizing at rest (M1)
(seconds) A1
Note: Award (M1)A0 for additional solutions to eg or .
[2 marks]
recognizing particle changes direction when OR when (M1)
A2
[3 marks]
distance travelled OR
(A1)
(metres) A1
[2 marks]
Examiners report
The majority of candidates found this question challenging but were often able to gain some of the marks in each part. However, it was not uncommon to see candidates manage either all of this question, or none of it, which unfortunately suggested that not all candidates had covered this content.
In part (a), while many recognized when the particle is at rest, a common error was to assume that . It was pleasing to see the majority of those who had the correct equation, manage to progress to the correct value of t, having recognised the domain and the angle measure. Inevitably, a few candidates ignored the domain and obtained , or found .
Part (b) was not well done. The most successful approach was to use the GDC to find the gradient of the curve at the value of t obtained in part (a). This was well communicated, concise and generally accurate, although some either rounded incorrectly, or obtained rather than . Of those that did not use the GDC, many were aware that and made an attempt to find an expression for . However, the majority appeared not to recognise when the particle would change direction and went on to substitute an incorrect value of , not realising that they had obtained the required value earlier.
Those that had been successful in parts (a) and (b), particularly if they had been using their GDC, were generally able to complete part (c). However, it was disappointing that many candidates who understood that an integral was required, did not refer to the formula booklet and omitted the absolute value from the integrand.
Question
A particle moves in a straight line such that its velocity, , at time seconds is given by .
Determine when the particle changes its direction of motion.
Find the times when the particle’s acceleration is .
Find the particle’s acceleration when its speed is at its greatest.
Markscheme
recognises the need to find the value of when (M1)
(s) A1
[2 marks]
recognises that (M1)
(s) A1A1
Note: Award M1A1A0 if the two correct answers are given with additional values outside .
[3 marks]
speed is greatest at (A1)
A1
[2 marks]
Examiners report
In part (a) many did not realize the change of motion occurred when . A common error was finding or thinking that it was at the maximum of .
In part (b), most candidates knew to differentiate but some tried to substitute in for , while others struggled to differentiate the function by hand rather than using the GDC. Many candidates tried to solve the equation analytically and did not use their technology. Of those who did, many had their calculators in degree mode.
Almost all candidates who attempted part (c) thought the greatest speed was the same as the maximum of .
Question
All lengths in this question are in centimetres.
A solid metal ornament is in the shape of a right pyramid, with vertex and square base . The centre of the base is . Point has coordinates and point has coordinates .
The volume of the pyramid is , correct to three significant figures.
Find .
Given that , find .
Find the height of the pyramid, .
A second ornament is in the shape of a cuboid with a rectangular base of length , width and height . The cuboid has the same volume as the pyramid.
The cuboid has a minimum surface area of . Find the value of .
Markscheme
attempt to use the distance formula to find (M1)
A1
[2 marks]
METHOD 1
attempt to apply cosine rule OR sine rule to find (M1)
OR (A1)
A1
METHOD 2
Let be the midpoint of
attempt to apply right-angled trigonometry on triangle (M1)
(A1)
A1
[3 marks]
METHOD 1
equating volume of pyramid formula to (M1)
(A1)
A1
METHOD 2
Let be the midpoint of
(M1)
(A1)
A1
[3 marks]
(A1)
A1
Note: Condone use of .
attempt to substitute into their expression for surface area (M1)
EITHER
attempt to find minimum turning point on graph of area function (M1)
OR
OR (M1)
THEN
minimum surface area A1
[5 marks]
Examiners report
Parts (a), (b) and (c) were completed well by many candidates, but few were able to make any significant progress in part (d).
In part (a), many candidates were able to apply the distance formula and successfully find AV. However, a common error was to work in two-dimensions and to apply Pythagoras' Theorem once, neglecting completely the z-coordinates. Many recognised the need to use either the sine or cosine rule in part (b) to find the length AB. Common errors in this part included: the GDC being set incorrectly in radians; applying right-angled trigonometry on a 40°, 40°, 70° triangle; or using as the length of AX in triangle AVX.
Despite the formula for the volume of a pyramid being in the formula booklet, a common error in part (c) was to omit the factor of from the volume formula, or not to recognise that the area of the base of the pyramid was .
The most challenging part of this question proved to be the optimization of the surface area of the cuboid in part (d). Although some candidates were able to form an equation involving the volume of the cuboid, an expression for the surface area eluded most. A common error was to gain a surface area which involved eight sides rather than six. It was surprising that few who were able to find both the equation and an expression were able to progress any further. Of those that did, few used their GDC to find the minimum surface area directly, with most preferring the more time consuming analytical approach.
Question
A scientist conducted a nine-week experiment on two plants, and , of the same species. He wanted to determine the effect of using a new plant fertilizer. Plant was given fertilizer regularly, while Plant was not.
The scientist found that the height of Plant , at time weeks can be modelled by the function , where .
The scientist found that the height of Plant , at time weeks can be modelled by the function , where .
Use the scientist’s models to find the initial height of
Plant .
Plant correct to three significant figures.
Find the values of when .
For , find the total amount of time when the rate of growth of Plant was greater than the rate of growth of Plant .
Markscheme
(cm) A1
[1 mark]
(M1)
(cm) A1
[2 marks]
attempts to solve for (M1)
(weeks) A2
[3 marks]
recognises that and are required (M1)
attempts to solve for (M1)
and OR and OR and (A1)
Note: Award full marks for .
Award subsequent marks for correct use of these exact values.
OR OR
(A1)
attempts to calculate the total amount of time (M1)
(weeks) A1
[6 marks]
Examiners report
Many students did not change their calculators back to radian mode. This meant they had no chance of correctly answering parts (c) and (d), since even if follow through was given, there were not enough intersections on the graphs.
Most managed part (a) and some attempted to equate the functions in part b) but few recognised that 'rate of growth' was the derivatives of the given functions, and of those who did, most were unable to find them.
Almost all the candidates who did solve part (c) gave the answer , when working with more significant figures would have given them 3.14. They lost the last mark.
Question
The function is defined by , where .
For the graph of
The graphs of and intersect at and , where .
write down the equation of the vertical asymptote.
find the equation of the horizontal asymptote.
Find .
Using an algebraic approach, show that the graph of is obtained by a reflection of the graph of in the -axis followed by a reflection in the -axis.
Find the value of and the value of .
Hence, find the area enclosed by the graph of and the graph of .
Markscheme
A1
[1 mark]
attempt to substitute into OR table with large values of OR sketch of showing asymptotic behaviour (M1)
A1
[2 marks]
attempt to interchange and (seen anywhere) M1
OR (A1)
OR (A1)
(accept ) A1
[4 marks]
reflection in -axis given by (M1)
(A1)
reflection of their in -axis given by accept "now " M1
OR A1
AG
Note: If the candidate attempts to show the result using a particular coordinate on the graph of rather than a general coordinate on the graph of , where appropriate, award marks as follows:
M0A0 for eg
M0A0 for
[4 marks]
attempt to solve using graph or algebraically (M1)
AND A1
Note: Award (M1)A0 if only one correct value seen.
[2 marks]
attempt to set up an integral to find area between and (M1)
(A1)
A1
[3 marks]
Examiners report
Candidates mostly found the first part of this question accessible, with many knowing how to find the equation of both asymptotes. Common errors included transposing the asymptotes, or finding where an asymptote occurred but not giving it as an equation.
Candidates knew how to start part (b)(i), with most attempting to find the inverse function by firstly interchanging and . However, many struggled with the algebra required to change the subject, and were not awarded all the marks. A common error in this part was for candidates to attempt to find an expression for , rather than one for . Few candidates were able to answer part (b)(ii). Many appeared not to know that a reflection in the -axis is given by , or that a reflection in the -axis is given by . Many of those that did, multiplied both the numerator and denominator by when taking the negative of their , i.e. was often simplified as . However, the majority of candidates either did not attempt this question part or attempted to describe a graphical approach often involving a specific point, rather than an algebraic approach.
Those that attempted part (c), and had the correct expression for , were usually able to gain all the marks. However, those that had an incorrect expression, or had found , often proceeded to find an area, even when there was not an area enclosed by their two curves.
Question
A particle moves in a straight line such that its velocity, , at time seconds is given by
.
The particle’s acceleration is zero at .
Find the value of .
Let be the distance travelled by the particle from to and let be the distance travelled by the particle from to .
Show that .
Markscheme
* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
attempts either graphical or symbolic means to find the value of when (M1)
A1
[2 marks]
attempts to find the value of either or (M1)
and A1A1
Note: Award as above for obtaining, for example, or
Note: Award a maximum of M1A1A0FT for use of an incorrect value of from part (a).
so AG
[3 marks]
Examiners report
Question
Consider the curves and for .
Find the -coordinates of the points of intersection of the two curves.
Find the area, , of the region enclosed by the two curves.
Markscheme
* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
attempts to solve (M1)
A1A1
Note: Award A1A0 if additional solutions outside the domain are given.
[3 marks]
(or equivalent) (M1)(A1)
Note: Award M1 for attempting to form an integrand involving “top curve” − “bottom curve”.
so A2
[4 marks]
Examiners report
Question
The displacement, in centimetres, of a particle from an origin, O, at time t seconds, is given by s(t) = t 2 cos t + 2t sin t, 0 ≤ t ≤ 5.
Find the maximum distance of the particle from O.
Find the acceleration of the particle at the instant it first changes direction.
Markscheme
use of a graph to find the coordinates of the local minimum (M1)
s = −16.513... (A1)
maximum distance is 16.5 cm (to the left of O) A1
[3 marks]
attempt to find time when particle changes direction eg considering the first maximum on the graph of s or the first t – intercept on the graph of s'. (M1)
t = 1.51986... (A1)
attempt to find the gradient of s' for their value of t, s" (1.51986...) (M1)
=–8.92 (cm/s2) A1
[4 marks]
